RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiatio

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	11
Exercise	11.5
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiatio

Question 1. Differentiate y = x^1/x with respect to x.

Solution:

We have,
=> y = x^1/x
On taking log of both the sides, we get,
=> log y = log x^1/x
=> log y = (1/x) (log x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(\frac{logx}{x})$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}(\frac{1}{x})+logx(\frac{-1}{x^2})$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}-\frac{logx}{x^2}$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{1-logx}{x^2}$
=> $\frac{dy}{dx}=\frac{(1-logx)y}{x^2}$
=> $\frac{dy}{dx}=\frac{(1-logx)x^{\frac{1}{x}}}{x^2}$

Question 2. Differentiate y = x^{sin x} with respect to x.

Solution:

We have,
=> y = x^{sin x}
On taking log of both the sides, we get,
=> log y = log x^{sin x}
=> log y = sin x log x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(sinxlogx)$
=> $\frac{1}{y}\frac{dy}{dx}=sinx(\frac{1}{x})+logx(cosx)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{sinx}{x}+logxcosx$
=> $\frac{dy}{dx}=y\left[\frac{sinx}{x}+logxcosx\right]$
=> $\frac{dy}{dx}=x^{sinx}\left[\frac{sinx}{x}+logxcosx\right]$

Question 3. Differentiate y = (1 + cos x)^x with respect to x.

Solution:

We have,
=> y = (1 + cos x)^x
On taking log of both the sides, we get,
=> log y = log (1 + cos x)^x
=> log y = x log (1 + cos x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (1 + cos x)]$
=> $\frac{1}{y}\frac{dy}{dx}=-xsinx(\frac{1}{1+cosx})+log(1+cosx)(1)$
=> $\frac{1}{y}\frac{dy}{dx}=-xsinx(\frac{1}{1+cosx})+log(1+cosx)(1)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{-xsinx}{1+cosx}+log(1+cosx)$
=> $\frac{dy}{dx}=y\left[log(1+cosx)-\frac{xsinx}{1+cosx}\right]$
=> $\frac{dy}{dx}=(1+cos x)^x\left[log(1+cosx)-\frac{xsinx}{1+cosx}\right]$

Question 4. Differentiate $y=x^{cos^{-1}x}$ with respect to x.

Solution:

We have,
=> $y=x^{cos^{-1}x}$
On taking log of both the sides, we get,
=> log y = log $x^{cos^{-1}x}$
=> log y = cos⁻¹ x log x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(cos^{−1} x log x)$
=> $\frac{1}{y}\frac{dy}{dx}=cos^{-1}x(\frac{1}{x})+logx(\frac{-1}{\sqrt{1-x^2}})$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}$
=> $\frac{dy}{dx}=y\left[\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}\right]$
=> $\frac{dy}{dx}=x^{cos^{-1}x}\left[\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}\right]$

Question 5. Differentiate y = (log x)^x with respect to x.

Solution:

We have,
=> y = (log x)^x
On taking log of both the sides, we get,
=> log y = log (log x)^x
=> log y = x log (log x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (log x)]$
=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{logx})(\frac{1}{x})+log(logx)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{logx}+log(logx)$
=> $\frac{dy}{dx}=y\left[\frac{1}{logx}+log(logx)\right]$
=> $\frac{dy}{dx}=(logx)^x\left[\frac{1}{logx}+log(logx)\right]$

Question 6. Differentiate y = (log x)^{cos x} with respect to x.

Solution:

We have,
=> y = (log x)^{cos x}
On taking log of both the sides, we get,
=> log y = log (log x)^{cos x}
=> log y = cos x log (log x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[cos x log (log x)]$
=> $\frac{1}{y}\frac{dy}{dx}=cosx(\frac{1}{logx})(\frac{1}{x})+log(logx)(-sinx)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{cosx}{xlogx}-sinxlog(logx)$
=> $\frac{dy}{dx}=y\left[\frac{cosx}{xlogx}-sinxlog(logx)\right]$
=> $\frac{dy}{dx}=(logx)^{cosx}\left[\frac{cosx}{xlogx}-sinxlog(logx)\right]$

Question 7. Differentiate y = (sin x)^{cos x} with respect to x.

Solution:

We have,
=> y = (sin x)^{cos x}
On taking log of both the sides, we get,
=> log y = log (sin x)^{cos x}
=> log y = cos x log (sin x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[cos x log (sin x)]$
=> $\frac{1}{y}\frac{dy}{dx}=cosx(\frac{1}{sinx})(cosx)+log(sinx)(-sinx)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{cos^2x}{sinx}-sinxlog(sinx)$
=> $\frac{1}{y}\frac{dy}{dx}=cotxcosx-sinxlog(sinx)$
=> $\frac{dy}{dx}=y\left[cotxcosx-sinxlog(sinx)\right]$
=> $\frac{dy}{dx}=(sinx)^{cosx}\left[cotxcosx-sinxlog(sinx)\right]$

Question 8. Differentiate y = e^{x log x} with respect to x.

Solution:

We have,
=> y=e^{x log x}
=> y = $e^{logx^x}$
=> y = x^x
On taking log of both the sides, we get,
=> log y = log x^x
=> log y = x log x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x log x)$
=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{x})+logx$
=> $\frac{1}{y}\frac{dy}{dx}=1+logx$
=> $\frac{dy}{dx}=y(1+logx)$
=> $\frac{dy}{dx}=x^{x}(1+logx)$

Question 9. Differentiate y = (sin x)^{log x} with respect to x.

Solution:

We have,
=> y = (sin x)^{log x}
On taking log of both the sides, we get,
=> log y = log (sin x)^{log x}
=> log y = log x log (sin x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log x log (sin x)]$
=> $\frac{1}{y}\frac{dy}{dx}=logx(\frac{1}{sinx})(cosx)+log(sinx)(\frac{1}{x})$
=> $\frac{1}{y}\frac{dy}{dx}=logxcotx+\frac{log(sinx)}{x}$
=> $\frac{dy}{dx}=y\left[logxcotx+\frac{log(sinx)}{x}\right]$
=> $\frac{dy}{dx}=(sinx)^{logx}\left[logxcotx+\frac{log(sinx)}{x}\right]$

Question 10. Differentiate y = 10^{log sin x} with respect to x.

Solution:

We have,
=> y = 10^{log sin x}
On taking log of both the sides, we get,
=> log y = log 10^{log sin x}
=> log y = log (sin x) log 10
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log (sin x) log 10]$
=> $\frac{1}{y}\frac{dy}{dx}=log10\frac{d}{dx}[log(sinx)]$
=> $\frac{1}{y}\frac{dy}{dx}=log10(\frac{1}{sinx})(cosx)$
=> $\frac{1}{y}\frac{dy}{dx}=log10cotx$
=> $\frac{dy}{dx}=ylog10cotx$
=> $\frac{dy}{dx}=10^{logsinx}[log10cotx]$

Question 11. Differentiate y = (log x)^{log x} with respect to x.

Solution:

We have,
=> y = (log x)^{log x}
On taking log of both the sides, we get,
=> log y = log (log x)^{log x}
=> log y = log x log (log x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log x log (log x)]$
=> $\frac{1}{y}\frac{dy}{dx}=logx(\frac{1}{logx})(\frac{1}{x})+log(logx)(\frac{1}{x})$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{log(logx)}{x}$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{1+log(logx)}{x}$
=> $\frac{dy}{dx}=\frac{y[1+log(logx)]}{x}$
=> $\frac{dy}{dx}=\frac{(logx)^{logx}[1+log(logx)]}{x}$

Question 12. Differentiate $y = 10^{10^x}$ with respect to x.

Solution:

We have,
=> $y = 10^{10^x}$
On taking log of both the sides, we get,
=> log y = log $10^{10^x}$
=> log y = 10^x log 10
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(10^x log 10)$
=> $\frac{1}{y}\frac{dy}{dx}=log10(10^xlog10)$
=> $\frac{1}{y}\frac{dy}{dx}=10^x(log10)^2$
=> $\frac{dy}{dx}=y10^x(log10)^2$
=> $\frac{dy}{dx}=10^{10^x}\left[10^x(log10)^2\right]$
=> $\frac{dy}{dx}=(10^{10^x+x})(log10)^2$

Question 13. Differentiate y = sin x^x with respect to x.

Solution:

We have,
=> y = sin x^x
=> sin⁻¹ y = x^x
On taking log of both the sides, we get,
=> log (sin⁻¹ y) = log x^x
=> log (sin⁻¹ y) = x log x
On differentiating both sides with respect to x, we get,
=> $(\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=\frac{d}{dx}(xlogx)$
=> $(\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=x(\frac{1}{x})+logx$
=> $(\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=1+logx$
=> $\frac{dy}{dx}=(1+logx)(sin^{-1}y)(\sqrt{1-y^2})$
=> $\frac{dy}{dx}=(1+logx)(sin^{-1}(sinx^x))(\sqrt{1-(sinx^x)^2})$
=> $\frac{dy}{dx}=(1+logx)(x^x)(\sqrt{1-sin^2x^x})$
=> $\frac{dy}{dx}=(1+logx)(x^x)(\sqrt{cos^2x^x})$
=> $\frac{dy}{dx}=x^xcosx^x(1+logx)$

Question 14. Differentiate y = (sin⁻¹x)^x with respect to x.

Solution:

We have,
=> y = (sin⁻¹x)^x
On taking log of both the sides, we get,
=> log y = (sin⁻¹x)^x
=> log y = x log (sin⁻¹x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (sin^{−1}x)]$
=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{sin^{-1}x})(\frac{1}{\sqrt{1-x^2}})+log (sin^{−1}x)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)$
=> $\frac{dy}{dx}=y\left[\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)\right]$
=> $\frac{dy}{dx}=(sin^{-1}x)^x\left[\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)\right]$

Question 15. Differentiate $y=x^{sin^{-1}x}$ with respect to x.

Solution:

We have,
=> $y=x^{sin^{-1}x}$
On taking log of both the sides, we get,
=> log y = log $x^{sin^{-1}x}$
=> log y = sin⁻¹x log x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(sin^{−1}x log x)$
=> $\frac{1}{y}\frac{dy}{dx}=sin^{−1}x(\frac{1}{x})+logx(\frac{1}{\sqrt{1-x^2}})$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}$
=> $\frac{dy}{dx}=y\left[\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}\right]$
=> $\frac{dy}{dx}=x^{sin^{-1}x}\left[\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}\right]$

Question 16. Differentiate $y=(tanx)^{\frac{1}{x}}$ with respect to x.

Solution:

We have,
=> $y=(tanx)^{\frac{1}{x}}$
On taking log of both the sides, we get,
=> log y = log $(tanx)^{\frac{1}{x}}$
=> log y = $\frac{1}{x}log(tanx)$
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[\frac{1}{x}log(tanx)]$
=> $\frac{1}{y}\frac{dy}{dx}=(\frac{1}{x})(\frac{1}{tanx})(sec^2x)+log(tanx)(\frac{-1}{x^2})$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}$
=> $\frac{dy}{dx}=y\left[\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}\right]$
=> $\frac{dy}{dx}=(tanx)^{\frac{1}{x}}\left[\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}\right]$

Question 17. Differentiate $y=x^{tan^{-1}x}$ with respect to x.

Solution:

We have,
=> $y=x^{tan^{-1}x}$
On taking log of both the sides, we get,
=> log y = log $y=x^{tan^{-1}x}$
=> log y = tan⁻¹ x log x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(tan^{−1} x log x)$
=> $\frac{1}{y}\frac{dy}{dx}=tan^{−1}x(\frac{1}{x})+logx(\frac{1}{1+x^2})$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}$
=> $\frac{dy}{dx}=y\left[\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}\right]$
=> $\frac{dy}{dx}=x^{tan^{-1}x}\left[\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}\right]$

Question 18. Differentiate the following with respect to x.

(i) y = x^x √x

Solution:

We have,
=> y = x^x √x
On taking log of both the sides, we get,
=> log y = log (x^x √x)
=> log y = log x^x + log √x
=> log y = x log x + $\frac{1}{2}logx$
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x log x +\frac{1}{2}logx)$
=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{x})+logx+(\frac{1}{2})(\frac{1}{x})$
=> $\frac{1}{y}\frac{dy}{dx}=1+logx+\frac{1}{2x}$
=> $\frac{dy}{dx}=y\left(1+logx+\frac{1}{2x}\right)$
=> $\frac{dy}{dx}=x^x\sqrt{x}\left(1+logx+\frac{1}{2x}\right)$

(ii) $y=x^{sinx-cosx}+\frac{x^2-1}{x^2+1}$

Solution:

We have,
=> $y=x^{sinx-cosx}+\frac{x^2-1}{x^2+1}$
=> $y=e^{logx^{sinx-cosx}}+\frac{x^2-1}{x^2+1}$
=> $y=e^{(sinx-cosx)logx}+\frac{x^2-1}{x^2+1}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}(e^{(sinx-cosx)logx}+\frac{x^2-1}{x^2+1})$
=> $\frac{dy}{dx}=(e^{(sinx-cosx)logx})[(sinx-cosx)\frac{1}{x}+logx(cosx+sinx)]+\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}$
=> $\frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{2x(x^2+1-x^2+1)}{(x^2+1)^2}$
=> $\frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{2x(2)}{(x^2+1)^2}$
=> $\frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{4x}{(x^2+1)^2}$

(iii) $y=x^{xcosx}+\frac{x^2+1}{x^2-1}$

Solution:

We have,
=> $y=x^{xcosx}+\frac{x^2+1}{x^2-1}$
=> $y=e^{logx^{xcosx}}+\frac{x^2+1}{x^2-1}$
=> $y=e^{xcosxlogx}+\frac{x^2+1}{x^2-1}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=(e^{xcosxlogx})[x((-sinx)logx+cosx(\frac{1}{x}))+cosxlogx]+\frac{(x^2-1)(2x)-(x^2+1)(2x)}{(x^2-1)^2}$
=> $\frac{dy}{dx}=(e^{xcosxlogx})[-xsinxlogx+cosx+cosxlogx]+\frac{2x(x^2-1-x^2-1)}{(x^2-1)^2}$
=> $\frac{dy}{dx}=(e^{xcosxlogx})[-xsinxlogx+cosx(1+logx)]+\frac{2x(-2)}{(x^2-1)^2}$
=> $\frac{dy}{dx}=(e^{xcosxlogx})[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}$
=> $\frac{dy}{dx}=x^{xcosx}[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}$

(iv) y = (x cos x)^x + (x sin x)^1/x

Solution:

We have,
=> y=(x cos x)^x + (x sin x)^1/x
=> $y=e^{log(x cos x)^x}+ e^{log(x sin x)^{\frac{1}{x}}}$
=> $y=e^{xlog(x cos x)}+ e^{\frac{1}{x}log(x sin x)}$
=> $y=e^{x(logx+logcosx)}+ e^{\frac{1}{x}(logx+logsin x)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=(e^{x(logx+logcosx)})[x(\frac{1}{x}+(\frac{1}{cosx})(-sinx))+log(xcosx)(1)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1}{x}(\frac{1}{x}+(\frac{1}{sinx})(cosx))+log(xsinx)(\frac{-1}{x^2})]$
=> $\frac{dy}{dx}=(e^{x(logx+logcosx)})[1-xtanx+log(xcosx)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1}{x^2}+\frac{1}{xcotx}-\frac{log(xsinx)}{x^2}]$
=> $\frac{dy}{dx}=(e^{x(logx+logcosx)})[1-xtanx+log(xcosx)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1-log(xsinx)+xcotx}{x^2}]$
=> $\frac{dy}{dx}=(xcosx)^x[1-xtanx+log(xcosx)]+ (xsinx)^{\frac{1}{x}}[\frac{1-log(xsinx)+xcotx}{x^2}]$

(v) $y=(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}$

Solution:

We have,
=> $y=(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}$
=> $y=e^{log(x+\frac{1}{x})^x}+e^{logx^{(1+\frac{1}{x})}}$
=> $y=e^{xlog(x+\frac{1}{x})}+e^{(1+\frac{1}{x})logx}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=(e^{xlog(x+\frac{1}{x})})[x(\frac{1}{x+\frac{1}{x}})(1-\frac{1}{x^2})+log(x+\frac{1}{x})]+(e^{(1+\frac{1}{x})logx})[(1+\frac{1}{x})(\frac{1}{x})+logx(\frac{-1}{x^2})]$
=> $\frac{dy}{dx}=(e^{xlog(x+\frac{1}{x})})[(\frac{x-\frac{1}{x}}{x+\frac{1}{x}})+log(x+\frac{1}{x})]+(e^{(1+\frac{1}{x})logx})[\frac{1}{x}+\frac{1}{x^2}-logx(\frac{1}{x^2})]$
=> $\frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}[\frac{1}{x}+\frac{1}{x^2}-logx(\frac{1}{x^2})]$
=> $\frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}[\frac{1}{x}+\frac{1}{x^2}-\frac{logx}{x^2}]$
=> $\frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}(\frac{x+1-logx}{x^2})$

(vi) y = e^{sin x} + (tan x)^x

Solution:

We have,
=> y = e^{sin x}+ (tan x)^x
=> $y = e^{sin x} + e^{log(tanx)^x}$
=> $y = e^{sin x} + e^{xlog(tanx)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}(e^{sin x} + e^{xlog(tanx)})$
=> $\frac{dy}{dx}=(e^{sin x})(cosx) + (e^{xlogtanx})\left[x(\frac{1}{tanx})(sec^2x)+log(tanx)\right]$
=> $\frac{dy}{dx}=e^{sin x}cosx+(tanx)^x\left[\frac{xsec^2x}{tanx}+log(tanx)\right]$

(vii) y = (cos x)^x + (sin x)^1/x

Solution:

We have,
=> y = (cos x)^x + (sin x)^1/x
=> $y = e^{log(cos x)^x} + e^{log(sin x)^{1/x}}$
=> $y = e^{xlog(cos x)} + e^{\frac{1}{x}log(sin x)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=(e^{xlog(cos x)})[x(\frac{1}{cosx})(-sinx)+log(cosx)] + (e^{\frac{1}{x}log(sin x)})[\frac{1}{x}(\frac{1}{sinx}(cosx))+log(sinx)(-\frac{1}{x^2})]$
=> $\frac{dy}{dx}=(e^{xlog(cos x)})[-xtanx+log(cosx)] + (e^{\frac{1}{x}log(sin x)})[\frac{cotx}{x}-\frac{log(sinx)}{x^2}]$
=> $\frac{dy}{dx}=(cosx)^x[-xtanx+log(cosx)] + (sinx)^{\frac{1}{x}}[\frac{cotx}{x}-\frac{log(sinx)}{x^2}]$

(viii) $y=x^{x^2-3}+(x-3)^{x^2}$ , for x > 3

Solution:

We have,
=> $y=x^{x^2-3}+(x-3)^{x^2}$
=> $y=e^{logx^{x^2-3}}+e^{log(x-3)^{x^2}}$
=> $y=e^{(x^2-3)logx}+e^{x^2log(x-3)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=(e^{(x^2-3)logx})[(x^2-3)(\frac{1}{x})+logx(2x)]+(e^{x^2log(x-3)})[x^2(\frac{1}{x-3})+log(x-3)(2x)]$
=> $\frac{dy}{dx}=(e^{(x^2-3)logx})[\frac{x^2-3}{x}+2xlogx]+(e^{x^2log(x-3)})[\frac{x^2}{x-3}+2xlog(x-3)]$
=> $\frac{dy}{dx}=x^{x^2-3}[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$

Question 19. Find dy/dx when y = e^x + 10^x + x^x.

Solution:

We have,
=> y = e^x + 10^x + x^x
=> $y = e^x + 10^x + e^{logx^x}$
=> $y = e^x + 10^x + e^{xlogx}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}(e^x + 10^x + e^{xlogx})$
=> $\frac{dy}{dx}=e^x+10^xlog10+e^{xlogx}[x(\frac{1}{x})+logx]$
=> $\frac{dy}{dx}=e^x+10^xlog10+x^x(1+logx)$
=> $\frac{dy}{dx}=e^x+10^xlog10+x^x(logex)$

Question 20. Find dy/dx when y = xⁿ + n^x+ x^x + nⁿ.

Solution:

We have,
=> y = xⁿ + n^x + x^x + nⁿ
=> $y=x^n + n^x + e^{logx^x} + n^n$
=> $y=x^n + n^x + e^{xlogx} + n^n$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}(x^n+n^x+e^{xlogx}+n^n)$
=> $\frac{dy}{dx}=nx^{n-1}+n^xlogn+e^{xlogx}[x(\frac{1}{x})+logx]+0$
=> $\frac{dy}{dx}=nx^{n-1}+n^xlogn+x^x(1+logx)$
=> $\frac{dy}{dx}=nx^{n-1}+n^xlogn+x^x(logex)$

Question 21. Find dy/dx when $y=\frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}$ .

Solution:

We have,
=> $y=\frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}$
=> $y=\frac{(x^2-1)^3(2x-1)}{(x-3)^{\frac{1}{2}}(4x-1)^{\frac{1}{2}}}$
On taking log of both the sides, we get,
=> $log y = log \frac{(x^2-1)^3(2x-1)}{(x-3)^{\frac{1}{2}}(4x-1)^{\frac{1}{2}}}$
=> $log y = log(x^2-1)^{3}+log(2x-1)-log(x-3)^{\frac{1}{2}}-log(4x-1)^{\frac{1}{2}}$
=> $log y = 3log(x^2-1)+log(2x-1)-\frac{1}{2}log(x-3)-\frac{1}{2}log(4x-1)$
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[3log(x^2-1)+log(2x-1)-\frac{1}{2}log(x-3)-\frac{1}{2}log(4x-1)]$
=> $\frac{1}{y}\frac{dy}{dx}=3(\frac{1}{x^2-1})(2x)+2(\frac{1}{2x-1})-\frac{1}{2}(\frac{1}{x-3})-\frac{1}{2}(\frac{1}{4x-1})(4)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2(x-3)}-\frac{2}{4x-1}$
=> $\frac{dy}{dx}=y\left[\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2(x-3)}-\frac{2}{4x-1}\right]$
=> $\frac{dy}{dx}=\frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}\left[\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2(x-3)}-\frac{2}{4x-1}\right]$

Question 22. Find dy/dx when $y=\frac{e^{ax}secxlogx}{\sqrt{1-2x}}$ .

Solution:

We have,
=> $y=\frac{e^{ax}secxlogx}{\sqrt{1-2x}}$
=> $y=\frac{e^{ax}secxlogx}{(1-2x)^{\frac{1}{2}}}$
On taking log of both the sides, we get,
=> $log y = log \frac{e^{ax}secxlogx}{(1-2x)^{\frac{1}{2}}}$
=> $log y = loge^{ax}+logsecx+log(logx)-\frac{1}{2}log(1-2x)$
=> $log y = ax+logsecx+log(logx)-\frac{1}{2}log(1-2x)$
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[ax+logsecx+log(logx)-\frac{1}{2}log(1-2x)]$
=> $\frac{1}{y}\frac{dy}{dx}=a+\frac{1}{secx}(secxtanx)+(\frac{1}{logx})(\frac{1}{x})-(\frac{1}{2})(\frac{1}{1-2x})(-2)$
=> $\frac{1}{y}\frac{dy}{dx}=a+tanx+\frac{1}{xlogx}+\frac{1}{1-2x}$
=> $\frac{dy}{dx}=y\left[a+tanx+\frac{1}{xlogx}+\frac{1}{1-2x}\right]$
=> $\frac{dy}{dx}=\frac{e^{ax}secxlogx}{\sqrt{1-2x}}\left[a+tanx+\frac{1}{xlogx}+\frac{1}{1-2x}\right]$

Question 23. Find dy/dx when y = e^3x sin 4x 2^x.

Solution:

We have
=> y = e^3x sin 4x 2^x.
On taking log of both the sides, we get,
=> log y = log (e^3x sin 4x 2^x)
=> log y = log e^3x + log (sin 4x) + log 2^x
=> log y = 3x log e + log (sin 4x) + x log 2
=> log y = 3x + log (sin 4x) + x log 2
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[3x + log (sin 4x) + x log 2]$
=> $\frac{1}{y}\frac{dy}{dx}=3+(\frac{1}{sin 4x})(4cos4x) + log2$
=> $\frac{1}{y}\frac{dy}{dx}=3+4cotx+log2$
=> $\frac{dy}{dx}=y(3+4cotx+log2)$
=> $\frac{dy}{dx}=2^xe^{3x}sin4x(3+4cotx+log2)$

Question 24. Find dy/dx when y = sin x sin 2x sin 3x sin 4x.

Solution:

We have,
=> y = sin x sin 2x sin 3x sin 4x
On taking log of both the sides, we get,
=> log y = log (sin x sin 2x sin 3x sin 4x)
=> log y = log sin x + log sin 2x + log sin 3x + log sin 4x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(log sin x + log sin 2x + log sin 3x + log sin 4x)$
=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{sinx}(cosx)+\frac{1}{sin2x}(2cos2x)+\frac{1}{sin3x}(3cos3x)+\frac{1}{sin4x}(4cos4x)$
=> $\frac{1}{y}\frac{dy}{dx}$ = cotx + 2cot2x + 3cot3x + 4cot4x
=> $\frac{dy}{dx}$ = y(cotx + 2cot2x + 3cot3x + 4cot4x)
=> $\frac{dy}{dx}$ = (sinxsin2x sin3xsin4x)(cotx + 2cot2x + 3cot3x + 4cot4x)

Question 25. Find dy/dx when y = x^{sin x} + (sin x)^x.

Solution:

We have,
=> y = x^{sin x} + (sin x)^x.
Let u = x^{sin x} and v = (sin x)^x. Therefore, y = u + v.
Now, u = x^{sin x}
On taking log of both the sides, we get,
=> log u = log x^{sin x}
=> log u = sin x log x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(sin x log x)$
=> $\frac{1}{u}\frac{du}{dx}=sin x(\frac{1}{x})+logxcosx$
=> $\frac{1}{u}\frac{du}{dx}=\frac{sinx}{x}+logxcosx$
=> $\frac{du}{dx}=u\left[\frac{sinx}{x}+logxcosx\right]$
=> $\frac{du}{dx}=x^{sinx}\left[\frac{sinx}{x}+logxcosx\right]$
Also, v = (sin x)^x
On taking log of both the sides, we get,
=> log v = log (sin x)^x
=> log v = x log sin x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(x log sin x)$
=> $\frac{1}{v}\frac{dv}{dx}=x(\frac{1}{sinx})(cosx)+log(sinx)$
=> $\frac{1}{v}\frac{dv}{dx}=xcotx+log(sinx)$
=> $\frac{dv}{dx}=v[xcotx+log(sinx)]$
=> $\frac{dv}{dx}=(sinx)^x[xcotx+log(sinx)]$
Now we have, y = u + v.
=> $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
=> $\frac{dy}{dx}=x^{sinx}\left[\frac{sinx}{x}+logxcosx\right]+(sinx)^x[xcotx+log(sinx)]$

Question 26. Find dy/dx when y = (sin x)^{cos x} + (cos x)^{sin x}.

Solution:

We have,
=> y = (sin x)^{cos x} + (cos x)^{sin x}
=> $y=e^{log(sin x)^{cos x}} + e^{log(cos x)^{sin x}}$
=> $y=e^{cosxlog(sin x)} + e^{sinxlog(cos x)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}(e^{cosxlog(sin x)} + e^{sinxlog(cos x)})$
=> $\frac{dy}{dx}=(e^{cosxlog(sin x)})[cosx(\frac{1}{sinx})cosx+log(sinx)(-sinx)] + (e^{sinxlog(cos x)})[sinx(\frac{1}{cosx})(-sinx)+log(cosx)(cosx)]$
=> $\frac{dy}{dx}$ = (sinx)^cosx[cosxcotx – sinxlog(sinx)] + (cosx)^sinx[-tanxsinx + cosxlog(cosx)]
=> $\frac{dy}{dx}$ = (sinx)^cosx[cosxcotx – sinxlog(sinx)] + (cosx)^sinx[cosxlog(cosx) – tanxsinx]

Question 27. Find dy/dx when y = (tan x)^{cot x} + (cot x)^{tan x}.

Solution:

We have,
=> y = (tan x)^{cot x} + (cot x)^{tan x}
=> $y=e^{log(tanx)^{cot x}} + e^{log(cot x)^{tan x}}$
=> $y=e^{cotxlog(tanx)} + e^{tanxlog(cot x)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}(e^{cotxlog(tanx)} + e^{tanxlog(cot x)})$
=> $\frac{dy}{dx}=(e^{cotxlog(tanx)})[cotx(\frac{1}{tanx})(sec^2x)+log(tanx)(-cosec^2x)] + (e^{tanxlog(cot x)})[tanx(\frac{1}{cotx})(-cosec^2x)+log(cotx)(sec^2x)]$
=> $\frac{dy}{dx}=(tanx)^{cotx}[cot^2x(sec^2x)-log(tanx)(cosec^2x)] + (cotx)^{tanx}[tan^2x(-cosec^2x)+log(cotx)(sec^2x)]$
=> $\frac{dy}{dx}$ = (tanx)^cotx[cosec²x – log(tanx)(cosec²x)] + (cotx)^tanx[-sec²x + log(cotx)(sec²x)]
=> $\frac{dy}{dx}$ = (tanx)^cotx[cosec²x – cosec²xlog(tanx)] + (cotx)^tanx[sec²xlog(cotx) – sec²x]

Question 28. Find dy/dx when y = (sin x)^x + sin⁻¹√x.

Solution:

We have,
=> y = (sin x)^x + sin⁻¹ √x
=> $y=e^{log(sinx)^{x}} + sin^{-1}\sqrt{x}$
=> $y=e^{xlog(sinx)} + sin^{-1}\sqrt{x}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}[e^{xlog(sinx)} + sin^{-1}\sqrt{x}]$
=> $\frac{dy}{dx}=(e^{xlog(sinx)})[x(\frac{1}{sinx})cosx+log(sinx)]+(\frac{1}{\sqrt{1-x}})(\frac{1}{2\sqrt{x}})$
=> $\frac{dy}{dx}=(e^{xlog(sinx)})[xcotx+log(sinx)]+\frac{1}{2\sqrt{x-x^2}}$
=> $\frac{dy}{dx}=(sinx)^x[xcotx+log(sinx)]+\frac{1}{2\sqrt{x-x^2}}$

Question 29. Find dy/dx when

(i) y = x^{cos x} + (sin x)^{tan x}

Solution:

We have,
=> y = x^{cos x} + (sin x)^{tan x}
=> $y=e^{log(x)^{cosx}} + e^{log(sinx)^{tanx}}$
=> $y=e^{cosxlog(x)} + e^{tanxlog(sinx)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}[e^{cosxlog(x)} + e^{tanxlog(sinx)}]$
=> $\frac{dy}{dx}=(e^{cosxlogx})[cosx(\frac{1}{x})+logx(-sinx)] + (e^{tanxlog(sinx)})[tanx(\frac{1}{sinx})(cosx)+log(sinx)sec^2x]$
=> $\frac{dy}{dx}=x^{cosx}[\frac{cosx}{x}-logxsinx] +(sinx)^{tanx}[tanx(cotx)+log(sinx)sec^2x]$
=> $\frac{dy}{dx}=x^{cosx}[\frac{cosx}{x}-logxsinx] +(sinx)^{tanx}[1+sec^2xlog(sinx)]$

(ii) y = x^x + (sin x)^x

Solution:

We have,
=> y = x^x + (sin x)^x
=> $y=e^{logx^{x}} + e^{log(sinx)^{x}}$
=> $y=e^{xlogx} + e^{xlog(sinx)}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}[e^{xlogx} + e^{xlog(sinx)}]$
=> $\frac{dy}{dx}=(e^{xlogx})[x(\frac{1}{x})+logx] + (e^{xlog(sinx)})[x(\frac{1}{sinx})(cosx)+log(sinx)]$
=> $\frac{dy}{dx}=(e^{xlogx})[1+logx] + (e^{xlog(sinx)})[xcotx+log(sinx)]$
=> $\frac{dy}{dx}=x^x(1+logx) + (sinx)^x[xcotx+log(sinx)]$

Question 30. Find dy/dx when y = (tan x)^{log x} + cos² (π/4).

Solution:

We have,
=> y = (tan x)^{log x} + cos² (π/4)
=> $y=e^{log(tanx)^{logx}} +cos^2(\frac{π}{4})$
=> $y=e^{logxlog(tanx)} +cos^2(\frac{π}{4})$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}[e^{logxlog(tanx)} +cos^2(\frac{π}{4})]$
=> $\frac{dy}{dx}=(e^{logxlog(tanx)})[logx(\frac{1}{tanx})(sec^2x)+log(tanx)(\frac{1}{x})] +0$
=> $\frac{dy}{dx}=(e^{logxlog(tanx)})[\frac{logxsec^2x}{tanx}+\frac{log(tanx)}{x}]$
=> $\frac{dy}{dx}=(tanx)^{logx}[\frac{logxsec^2x}{tanx}+\frac{log(tanx)}{x}]$

Question 31. Find dy/dx when $y=x^x+x^{\frac{1}{x}}$ .

Solution:

We have,
=> $y=x^x+x^{\frac{1}{x}}$
=> $y=e^{logx^{x}} + e^{logx^{\frac{1}{x}}}$
=> $y=e^{xlogx} + e^{\frac{1}{x}logx}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{d}{dx}[e^{xlogx} + e^{\frac{1}{x}logx}]$
=> $\frac{dy}{dx}=(e^{xlogx})[x(\frac{1}{x})+logx] + (e^{\frac{1}{x}logx})[(\frac{1}{x})(\frac{1}{x})+logx(\frac{-1}{x^2})]$
=> $\frac{dy}{dx}=(e^{xlogx})(1+logx) + (e^{\frac{1}{x}logx})[\frac{1}{x^2}-\frac{logx}{x^2}]$
=> $\frac{dy}{dx}=(e^{xlogx})(1+logx) + (e^{\frac{1}{x}logx})(\frac{1-logx}{x^2})$
=> $\frac{dy}{dx}=x^x(1+logx) + x^{\frac{1}{x}}(\frac{1-logx}{x^2})$

Question 32. Find dy/dx when y = (log x)^x+ x^logx.

Solution:

We have,
=> y = (log x)^x+ x^logx
Let u = (log x)^x and v = x^logx. Therefore, y = u + v.
Now, u = (log x)^x
On taking log of both the sides, we get,
=> log u = log (log x)^x
=> log u = x log (log x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}[x log (log x)]$
=> $\frac{1}{u}\frac{du}{dx}=[x(\frac{1}{logx})(\frac{1}{x})+log(logx)]$
=> $\frac{1}{u}\frac{du}{dx}=\frac{1}{logx}+log(logx)$
=> $\frac{du}{dx}=u[\frac{1}{logx}+log(logx)]$
=> $\frac{du}{dx}=(logx)^x[\frac{1}{logx}+log(logx)]$
=> $\frac{du}{dx}=(logx)^x[\frac{1+logxlog(logx)}{logx}]$
=> $\frac{du}{dx}=(logx)^{x-1}[1+logxlog(logx)]$
Also, v = x^logx
On taking log of both the sides, we get,
=> log v = log x^logx
=> log v = log x (log x)
=> log v = (log x)²
On differentiating both sides with respect to x, we get,
=> $\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}[(log x)^2]$
=> $\frac{1}{v}\frac{dv}{dx}=2(logx)(\frac{1}{x})$
=> $\frac{dv}{dx}=v(\frac{2logx}{x})$
=> $\frac{dv}{dx}=x^{logx}(\frac{2logx}{x})$
=> $\frac{dv}{dx}=2logx.x^{logx-1}$
Now, y = u + v
=> $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
=> $\frac{dy}{dx}=(logx)^{x-1}[1+logxlog(logx)]+2logx.x^{logx-1}$

Question 33. If x¹³y⁷ = (x+y)²⁰, prove that $\frac{dy}{dx}=\frac{y}{x}$ .

Solution:

We have,
=> x¹³y⁷ = (x+y)²⁰
On taking log of both the sides, we get,
=> log x¹³y⁷= log (x+y)²⁰
=> log x¹³+ log y⁷= log (x+y)²⁰
=> 13 log x + 7 log y = 20 log (x+y)
On differentiating both sides with respect to x, we get,
=> $13(\frac{1}{x})+7(\frac{1}{y})(\frac{dy}{dx})=20(\frac{1}{x+y})(1+\frac{dy}{dx})$
=> $\frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}(1+\frac{dy}{dx})$
=> $\frac{13}{x}+\frac{7}{y}\frac{dy}{dx}=\frac{20}{x+y}+\frac{20}{x+y}\frac{dy}{dx}$
=> $\frac{7}{y}\frac{dy}{dx}-\frac{20}{x+y}\frac{dy}{dx}=\frac{20}{x+y}-\frac{13}{x}$
=> $\frac{dy}{dx}(\frac{7}{y}-\frac{20}{x+y})=\frac{20x-13x-13y}{x(x+y)}$
=> $\frac{dy}{dx}(\frac{7x+7y-20y}{y(x+y)})=\frac{7x-13y}{x(x+y)}$
=> $\frac{dy}{dx}(\frac{7x-13y}{y(x+y)})=\frac{7x-13y}{x(x+y)}$
=> $\frac{dy}{dx}=\frac{y}{x}$
Hence proved.

Question 34. If x¹⁶y⁹ = (x² + y)¹⁷, prove that $x\frac{dy}{dx}=2y$ .

Solution:

We have,
=> x¹⁶y⁹ = (x²+ y)¹⁷
On taking log of both the sides, we get,
=> log x¹⁶y⁹ = log (x² + y)¹⁷
=> log x¹⁶ + log y⁹ = log (x² +y)¹⁷
=> 16 log x + 9 log y = 17 log (x² + y)
On differentiating both sides with respect to x, we get,
=> $16(\frac{1}{x})+9(\frac{1}{y})(\frac{dy}{dx})=17(\frac{1}{x^2+y})(2x+\frac{dy}{dx})$
=> $\frac{16}{x}+\frac{9}{y}\frac{dy}{dx}=\frac{34x}{x^2+y}+\frac{17}{x^2+y}\frac{dy}{dx}$
=> $\frac{9}{y}\frac{dy}{dx}-\frac{17}{x^2+y}\frac{dy}{dx}=\frac{34x}{x^2+y}-\frac{16}{x}$
=> $(\frac{9}{y}-\frac{17}{x^2+y})\frac{dy}{dx}=\frac{34x^2-16x^2-16y}{x(x^2+y)}$
=> $(\frac{9x^2+9y-17y}{y(x^2+y)})\frac{dy}{dx}=\frac{34x^2-16x^2-16y}{x(x^2+y)}$
=> $(\frac{9x^2-8y}{y(x^2+y)})\frac{dy}{dx}=\frac{18x^2-16y}{x(x^2+y)}$
=> $(\frac{9x^2-8y}{y})\frac{dy}{dx}=\frac{2(9x^2-8y)}{x}$
=> $\frac{dy}{dx}=\frac{2y}{x}$
=> $x\frac{dy}{dx}=2y$
Hence proved.

Question 35. If y = sin x^x, prove that $\frac{dy}{dx}=cos(x^x)×x^x(1+logx)$ .

Solution:

We have,
=> y = sin x^x
Let u = x^x. Now y = sin u.
On taking log of both the sides, we get,
=> log u = log x^x
=> log u = x log x
On differentiating both sides with respect to x, we get,
=> $\frac{1}{u}\frac{du}{dx}=x(\frac{1}{x})+logx$
=> $\frac{1}{u}\frac{du}{dx}=1+logx$
=> $\frac{du}{dx}=u(1+logx)$
=> $\frac{du}{dx}=x^x(1+logx)$
Now, y = sin u
=> $\frac{dy}{dx}=cosu\frac{du}{dx}$
=> $\frac{dy}{dx}=cosu×x^x(1+logx)$
=> $\frac{dy}{dx}=cosx^x×x^x(1+logx)$
Hence proved.

Question 36. If x^x + y^x = 1, prove that $\frac{dy}{dx}=\frac{x^x(1+logx)+y^xlogy}{xy^{x-1}}$ .

Solution:

We have,
=> x^x + y^x = 1
=> $e^{logx^x} + e^{logy^x} = 1$
=> $e^{xlogx} + e^{xlogy} = 1$
On differentiating both sides with respect to x, we get,
=> $(e^{xlogx})[x(\frac{1}{x})+logx] + (e^{xlogy})[x(\frac{1}{y})(\frac{dy}{dx})+logy] = 0$
=> $(e^{xlogx})(1+logx) + (e^{xlogy})[(\frac{x}{y})(\frac{dy}{dx})+logy] = 0$
=> $x^x(1+logx) + y^x[(\frac{x}{y})(\frac{dy}{dx})+logy] = 0$
=> $x^x(1+logx) + xy^{x-1}\frac{dy}{dx}+y^xlogy = 0$
=> $xy^{x-1}\frac{dy}{dx} = -[x^x(1+logx)+y^xlogy ]$
=> $\frac{dy}{dx} = \frac{-[x^x(1+logx)+y^xlogy]}{xy^{x-1}}$
Hence proved.

Question 37. If x^y × y^x = 1, prove that $\frac{dy}{dx}=\frac{-y(y+xlogy)}{x(ylogx+x)}$ .

Solution:

We have,
=> x^y × y^x = 1
On taking log of both the sides, we get,
=> log (x^y× y^x) = log 1
=> log x^y+ log y^x = log 1
=> y log x + x log y = log 1
On differentiating both sides with respect to x, we get,
=> $y(\frac{1}{x})+logx(\frac{dy}{dx})+x(\frac{1}{y})(\frac{dy}{dx})+log y = 0$
=> $\frac{y}{x}+logx(\frac{dy}{dx})+(\frac{x}{y})(\frac{dy}{dx})+log y = 0$
=> $(\frac{y}{x}+logy)+(logx+\frac{x}{y})(\frac{dy}{dx}) = 0$
=> $(\frac{y+xlogy}{x})+(\frac{ylogx+x}{y})(\frac{dy}{dx}) = 0$
=> $(\frac{ylogx+x}{y})(\frac{dy}{dx})= -(\frac{y+xlogy}{x})$
=> $\frac{dy}{dx}=\frac{-y(y+xlogy)}{x(ylogx+x)}$
Hence proved.

Question 38. If x^y+ y^x = (x+y)^x+y, find dy/dx.

Solution:

We have,
=> x^y + y^x = (x+y)^x+y
=> $e^{logx^y} + e^{logy^x} = e^{log(x+y)^{x+y}}$
=> $e^{ylogx} + e^{xlogy} = e^{(x+y)log(x+y)}$
On differentiating both sides with respect to x, we get,
=> $(e^{ylogx})[y(\frac{1}{x})+logx(\frac{dy}{dx})] + (e^{xlogy})[x(\frac{1}{y})(\frac{dy}{dx})+logy]=(e^{(x+y)log(x+y)})[(x+y)(\frac{1}{x+y})(1+\frac{dy}{dx})+log(x+y)(1+\frac{dy}{dx})]$
=> $e^{ylogx}[\frac{y}{x}+logx(\frac{dy}{dx})] + (e^{xlogy})[(\frac{x}{y})(\frac{dy}{dx})+logy]=(e^{(x+y)log(x+y)})[1+\frac{dy}{dx}+log(x+y)(1+\frac{dy}{dx})]$
=> $x^y[\frac{y}{x}+logx(\frac{dy}{dx})] + y^x[(\frac{x}{y})(\frac{dy}{dx})+logy]=(x+y)^{x+y}[1+\frac{dy}{dx}+log(x+y)(1+\frac{dy}{dx})]$
=> $\frac{dy}{dx}[x^ylogx+xy^{x-1}-(x+y)^{x+y}(1+log(x+y))]=(x+y)^{x+y}(1+log(x+y))-yx^{y-1}-y^xlogy$
=> $\frac{dy}{dx}=\frac{(x+y)^{x+y}(1+log(x+y))-yx^{y-1}-y^xlogy}{x^ylogx+xy^{x-1}-(x+y)^{x+y}(1+log(x+y))}$

Question 39. If x^m yⁿ = 1, prove that $\frac{dy}{dx}=-\frac{my}{nx}$ .

Solution:

We have,
=> x^m yⁿ = 1
On taking log of both the sides, we get,
=> log (x^m yⁿ)= log 1
=> log x^m + log yⁿ = log 1
=> m log x + n log y = log 1
On differentiating both sides with respect to x, we get,
=> $m(\frac{1}{x})+n(\frac{1}{y})(\frac{dy}{dx}) = 0$
=> $\frac{m}{x}+\frac{n}{y}(\frac{dy}{dx}) = 0$
=> $\frac{n}{y}(\frac{dy}{dx}) = -\frac{m}{x}$
=> $\frac{dy}{dx}= -\frac{my}{xn}$
Hence proved.

Question 40. If y^x = e^y−x, prove that $\frac{dy}{dx}=\frac{(1+logy)^2}{logy}$ .

Solution:

We have,
=> y^x = e^y−x
On taking log of both the sides, we get,
=> log y^x = log e^y−x
=> x log y = (y − x) log e
=> x log y = y − x
On differentiating both sides with respect to x, we get,
=> $x(\frac{1}{y})(\frac{dy}{dx})+logy=\frac{dy}{dx}−1$
=> $\frac{x}{y}\frac{dy}{dx}+logy=\frac{dy}{dx}−1$
=> $(\frac{x}{y}-1)\frac{dy}{dx}=−1-logy$
=> $(\frac{x}{y}-1)\frac{dy}{dx}=−(1+logy)$
=> $(\frac{y}{1+logy})\frac{dy}{dx}=−(1+logy)$
=> $(\frac{1-1-logy}{1+logy})\frac{dy}{dx}=−(1+logy)$
=> $\frac{dy}{dx}=−\frac{(1+logy)^2}{-logy}$
=> $\frac{dy}{dx}=\frac{(1+logy)^2}{logy}$
Hence proved.
Question 41. If (sin x)^y = (cos y)^x, prove that $\frac{dy}{dx}=\frac{log cosy-ycotx}{logsinx+xtany}$ .
Solution:

We have,
=> (sin x)^y = (cos y)^x
On taking log of both the sides, we get,
=> log (sin x)^y = log (cos y)^x
=> y log (sin x) = x log (cos y)
On differentiating both sides with respect to x, we get,
=> $y[(\frac{1}{sinx})(cosx)]+log(sin x)(\frac{dy}{dx})=x[(\frac{1}{cosy})(-siny)(\frac{dy}{dx})]+log (cos y)$
=> $ycotx+log(sin x)(\frac{dy}{dx})=-xtany(\frac{dy}{dx})+log (cos y)$
=> $[log(sin x)+xtany](\frac{dy}{dx})=log (cos y)-ycotx$
=> $\frac{dy}{dx}=\frac{log cosy-ycotx}{logsinx+xtany}$
Hence proved.
Question 42. If (cos x)^y = (tan y)^x, prove that \frac{dy}{dx}=\frac{log tany+ytanx}{logcosx-xsecycosecy} .
Solution:
We have, (cos x)^y = (tan y)^x
On taking log of both the sides, we get,
=> log (cos x)^y = log (tan y)^x
=> y log (cos x) = x log (tan y)
On differentiating both sides with respect to x, we get,
=> $y[(\frac{1}{cosx})(-sinx)]+log (cos x)(\frac{dy}{dx})=x[(\frac{1}{tany})(sec^2y)(\frac{dy}{dx})]+log (tan y)$
=> $-ytanx+log(cos x)(\frac{dy}{dx})=(\frac{xsec^2y}{tany})(\frac{dy}{dx})+log (tan y)$
=> $-ytanx-log (tan y)=(\frac{xsec^2y}{tany})(\frac{dy}{dx})-log(cos x)(\frac{dy}{dx})$
=> $ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(\frac{xsec^2y}{tany})(\frac{dy}{dx})$
=> $ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(\frac{x}{cos^2y}×\frac{cosy}{siny})(\frac{dy}{dx})$
=> $ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(\frac{x}{cosy}×\frac{1}{siny})(\frac{dy}{dx})$
=> $ytanx+log (tan y)=log(cos x)(\frac{dy}{dx})-(xsecycosecy)(\frac{dy}{dx})$
=> $ytanx+log (tan y)=[log(cos x)-(xsecycosecy)]\frac{dy}{dx}$
=> $[log(cos x)-(xsecycosecy)]\frac{dy}{dx}=ytanx+log (tan y)$
=> $\frac{dy}{dx}=\frac{ytanx+log (tan y)}{log(cos x)-xsecycosecy}$
Hence proved.
Question 43. If e^x + e^y = e^x+y, prove that $\frac{dy}{dx}+e^{y-x}=0$ .
Solution:
We have,
=> e^x + e^y = e^x+y
On differentiating both sides with respect to x, we get,
=> $e^x+e^y(\frac{dy}{dx})=e^{x+y}(1+\frac{dy}{dx})$
=> $e^x+e^y(\frac{dy}{dx})=e^{x+y}+e^{x+y}(\frac{dy}{dx})$
=> $e^y(\frac{dy}{dx})-e^{x+y}(\frac{dy}{dx})=e^{x+y}-e^x$
=> $\frac{dy}{dx}(e^y-e^{x+y})=e^{x+y}-e^x$
=> $\frac{dy}{dx}=\frac{e^{x+y}-e^x}{e^y-e^{x+y}}$
=> $\frac{dy}{dx}=\frac{e^x+e^y-e^x}{e^y-e^x-e^y}$
=> $\frac{dy}{dx}=-e^{y-x}$
=> $\frac{dy}{dx}+e^{y-x}=0$
Hence proved.
Question 44. If e^y = y^x, prove that $\frac{dy}{dx}=\frac{(logy)^2}{logy-1}$ .
Solution:
We have,
=> e^y = y^x
On taking log of both the sides, we get,
=> log e^y = log y^x
=> y log e = x log y
=> y = x log y
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=x(\frac{1}{y})(\frac{dy}{dx})+logy$
=> $\frac{dy}{dx}=(\frac{x}{y})(\frac{dy}{dx})+logy$
=> $\frac{dy}{dx}-(\frac{x}{y})(\frac{dy}{dx})=logy$
=> $\frac{dy}{dx}(1-\frac{x}{y})=logy$
=> $\frac{dy}{dx}(\frac{y-x}{y})=logy$
=> $\frac{dy}{dx}=\frac{ylogy}{y-x}$
=> $\frac{dy}{dx}=\frac{ylogy}{y-\frac{y}{logy}}$
=> $\frac{dy}{dx}=\frac{ylogy}{\frac{ylogy-y}{logy}}$
=> $\frac{dy}{dx}=\frac{y(logy)^2}{ylogy-y}$
=> $\frac{dy}{dx}=\frac{y(logy)^2}{y(logy-1)}$
Hence proved.
Question 45. If e^x+y − x = 0, prove that $\frac{dy}{dx}=\frac{1-x}{x}$ .
Solution:
We have,
=> e^x+y − x = 0
On differentiating both sides with respect to x, we get,
=> $e^{x+y}(1+\frac{dy}{dx}) − 1 = 0$
=> $e^{x+y}(1+\frac{dy}{dx})= 1$
Now, we know e^x+y − x = 0
=> e^x+y = x
So, we get,
=> $e^{x+y}(1+\frac{dy}{dx})= 1$
=> $x(1+\frac{dy}{dx})= 1$
=> $1+\frac{dy}{dx}=\frac{1}{x}$
=> $\frac{dy}{dx}=\frac{1}{x}-1$
=> $\frac{dy}{dx}=\frac{1-x}{x}$
Hence proved.
Question 46. If y = x sin (a+y), prove that $\frac{dy}{dx}=\frac{sin^2(a+y)}{sin(a+y)-ycos(a+y)}$ .
Solution:
We have,
=> y = x sin (a+y)
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=x[cos(a+y)(\frac{dy}{dx})]+sin(a+y))(1)$
=> $\frac{dy}{dx}=xcos(a+y)(\frac{dy}{dx})+sin(a+y))$
=> $\frac{dy}{dx}-xcos(a+y)(\frac{dy}{dx})=sin(a+y))$
=> $\frac{dy}{dx}(1-xcos(a+y))=sin(a+y))$
Now we know, y = x sin (a+y)
=> $x=\frac{y}{sin(a+y)}$
So, we get,
=> $\frac{dy}{dx}(1-\frac{y}{sin(a+y)}cos(a+y))=sin(a+y))$
=> $\frac{dy}{dx}(1-\frac{ycos(a+y)}{sin(a+y)})=sin(a+y))$
=> $\frac{dy}{dx}(\frac{sin(a+y)-ycos(a+y)}{sin(a+y)})=sin(a+y))$
=> $\frac{dy}{dx}=\frac{sin^2(a+y)}{sin(a+y)-ycos(a+y)}$
Hence proved.
Question 47. If x sin (a+y) + sin a cos (a+y) = 0, prove that $\frac{dy}{dx}=\frac{sin^2(a+y)}{sina}$ .
Solution:
We have,
=> x sin (a+y) + sin a cos (a+y) = 0
On differentiating both sides with respect to x, we get,
=> $x(cos (a+y))(\frac{dy}{dx})+sin(a+y)(1)+sina [-sin(a+y)(\frac{dy}{dx})] = 0$
=> $x(cos (a+y))(\frac{dy}{dx})+sin(a+y)-sinasin(a+y)(\frac{dy}{dx}) = 0$
=> $\frac{dy}{dx}(xcos(a+y)-sinasin(a+y))+sin(a+y)= 0$
=> $\frac{dy}{dx}= \frac{-sin(a+y)}{xcos(a+y)-sinasin(a+y)}$
Now we know, x sin (a+y) + sin a cos (a+y) = 0
=> $x=\frac{-sinacos(a+y)}{sin(a+y)}$
So, we get,
=> $\frac{dy}{dx}= \frac{-sin(a+y)}{(\frac{-sinacos(a+y)}{sin(a+y)})cos(a+y)-sinasin(a+y)}$
=> $\frac{dy}{dx}= \frac{-sin(a+y)}{(\frac{-sinacos^2(a+y)}{sin(a+y)})-sinasin(a+y)}$
=> $\frac{dy}{dx}= \frac{-sin(a+y)}{(\frac{-sinacos^2(a+y)-sinasin^2(a+y)}{sin(a+y)})}$
=> $\frac{dy}{dx}= \frac{-sin(a+y)}{\frac{-sina(cos^2(a+y)+sin^2(a+y))}{sin(a+y)}}$
=> $\frac{dy}{dx}= \frac{-sin(a+y)}{\frac{-sina(1)}{sin(a+y)}}$
=> $\frac{dy}{dx}= \frac{sin(a+y)}{\frac{sina}{sin(a+y)}}$
=> $\frac{dy}{dx}= \frac{sin^2(a+y)}{sina}$
Hence proved.
Question 48. If (sin x)^y = x + y, prove that $\frac{dy}{dx}=\frac{1-(x+y)ycotx}{(x+y)logsinx-1}$ .
Solution:
We have,
=> (sin x)^y = x + y
On taking log of both the sides, we get,
=> log (sin x)^y = log (x + y)
=> y log sin x = log (x + y)
On differentiating both sides with respect to x, we get,
=> $y(\frac{1}{sinx})(cosx)+log sin x(\frac{dy}{dx})=\frac{1}{x + y}(1+\frac{dy}{dx})$
=> $ycotx+log sin x(\frac{dy}{dx})=\frac{1}{x + y}+\frac{1}{x+y}\frac{dy}{dx}$
=> $logsinx(\frac{dy}{dx})-\frac{1}{x+y}\frac{dy}{dx}=\frac{1}{x + y}-ycotx$
=> $\frac{dy}{dx}(logsinx-\frac{1}{x+y})=\frac{1}{x + y}-ycotx$
=> $\frac{dy}{dx}(\frac{(x+y)logsinx-1}{x+y})=\frac{1-y(x+y)cotx}{x + y}$
=> $\frac{dy}{dx}[(x+y)logsinx-1]=[1-y(x+y)cotx]$
=> $\frac{dy}{dx}=\frac{1-y(x+y)cotx}{(x+y)logsinx-1}$
Hence proved.
Question 49. If xy log (x+y) = 1, prove that $\frac{dy}{dx}=-\frac{y(x^2y+x+y)}{x(xy^2+x+y)}$ .
Solution:
We have,
=> xy log (x+y) = 1
On differentiating both sides with respect to x, we get,
=> $xy(\frac{1}{x+y})(1+\frac{dy}{dx})+log (x+y)(x\frac{dy}{dx}+y) = 0$
=> $\frac{xy}{x+y}(1+\frac{dy}{dx})+xlog(x+y)\frac{dy}{dx}+ylog(x+y)=0$
=> $\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+xlog(x+y)\frac{dy}{dx}+ylog(x+y)=0$
Now, we know, xy log (x+y) = 1.
=> $log (x+y) = \frac{1}{xy}$
So, we get,
=> $\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+xlog(x+y)\frac{dy}{dx}+ylog(x+y)=0$
=> $\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+x(\frac{1}{xy})\frac{dy}{dx}+y(\frac{1}{xy})=0$
=> $\frac{xy}{x+y}+\frac{xy}{x+y}\frac{dy}{dx}+\frac{1}{y}\frac{dy}{dx}+\frac{1}{x}=0$
=> $\frac{dy}{dx}(\frac{xy}{x+y}+\frac{1}{y})+\frac{xy}{x+y}+\frac{1}{x}=0$
=> $\frac{dy}{dx}(\frac{xy^2+x+y}{y(x+y)})+\frac{x^2y+x+y}{x(x+y)}=0$
=> $\frac{dy}{dx}(\frac{xy^2+x+y}{y(x+y)})=-\frac{x^2y+x+y}{x(x+y)}$
=> $\frac{dy}{dx}(\frac{xy^2+x+y}{y})=-\frac{x^2y+x+y}{x}$
=> $\frac{dy}{dx}=-\frac{y(x^2y+x+y)}{x(xy^2+x+y)}$
Hence proved.
Question 50. If y = x sin y, prove that $\frac{dy}{dx}=\frac{y}{x(1-xcosy)}$ .
Solution:
We have,
=> y = x sin y
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=x(cosy)(\frac{dy}{dx})+sin y(1)$
=> $\frac{dy}{dx}=xcosy\frac{dy}{dx}+sin y$
=> $\frac{dy}{dx}(1-xcosy)=sin y$
=> $\frac{dy}{dx}=\frac{sin y}{1-xcosy}$
Now, we know y = x sin y
=> $siny = \frac{y}{x}$
So, we get,
=> $\frac{dy}{dx}=\frac{sin y}{1-xcosy}$
=> $\frac{dy}{dx}=\frac{\frac{y}{x}}{1-xcosy}$
=> $\frac{dy}{dx}=\frac{y}{x(1-xcosy)}$
Hence proved.
Question 51. Find the derivative of the function f(x) given by,
f(x) = (1+x) (1+x²) (1+x⁴) (1+x⁸) and hence find f'(1).
Solution:
Here we are given,
=> f(x) = (1+x) (1+x²) (1+x⁴) (1+x⁸)
On differentiating both sides with respect to x, we get,
=> $f'(x)=(1+x)(1+x^2)\frac{d}{dx}(1+x^8)+(1+x)(1+x^2)(1+x^8)\frac{d}{dx}(1+x^4)+(1+x)(1+x^4)(1+x^8)\frac{d}{dx}(1+x^2)+(1+x^2)(1+x^4)(1+x^8)\frac{d}{dx}(1+x)$
=>
Now, the value of f'(x) at 1 is,
=> f'(1) = (1 + 1) (1 + 1) (1 + 1) (8) + (1 + 1) (1 + 1) (1 + 1) (4) + (1 + 1) (1 + 1) (1 + 1) (2) + (1 + 1) (1 + 1) (1 + 1) (1)
=> f'(1) = (2) (2) (2) (8) + (2) (2) (2) (4) + (2) (2) (2) (2) + (2) (2) (2) (1)
=> f'(1) = 64 + 32 + 16 + 8
=> f'(1) = 120
Therefore, the value of f'(1) is 120.
Question 52. If $y=log\frac{x^2+x+1}{x^2-x+1}+\frac{2}{\sqrt{3}}tan^{-1}(\frac{\sqrt{3}x}{1-x^2})$ , find $\frac{dy}{dx}$ .
Solution:
We are given,
=> $y=log\frac{x^2+x+1}{x^2-x+1}+\frac{2}{\sqrt{3}}tan^{-1}(\frac{\sqrt{3}x}{1-x^2})$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{1}{\frac{x^2+x+1}{x^2-x+1}}\frac{d}{dx}(\frac{x^2+x+1}{x^2-x+1})+\frac{2}{\sqrt{3}}\left[\frac{1}{1+(\frac{\sqrt{3}x}{1-x^2})^2}\right]\frac{d}{dx}(\frac{\sqrt{3}x}{1-x^2})$
=> $\frac{dy}{dx}=\frac{x^2-x+1}{x^2+x+1}\left[\frac{(x^2-x+1)(2x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}\right]+\frac{2}{\sqrt{3}}\left[\frac{1}{1+\frac{3x^2}{(1-x^2)^2}}\right]\left[\frac{(1-x^2)(\sqrt{3})-\sqrt{3}x(-2x)}{(1-x^2)^2}\right]$
=> $\frac{dy}{dx}=\frac{1}{x^2+x+1}\left[\frac{2x^3+x^2-2x^2-x+2x+1-2x^3-2x^2-2x+x^2+x+1}{x^2-x+1}\right]+\frac{2}{\sqrt{3}}\left[\frac{1}{\frac{(1-x^2)^2+3x^2}{(1-x^2)^2}}\right]\left[\frac{\sqrt{3}-\sqrt{3}x^2+2\sqrt{3}x^2}{(1-x^2)^2}\right]$
=> $\frac{dy}{dx}=\frac{-2x^2+2}{x^4+2x^2+1-x^2}+\frac{2}{\sqrt{3}}\left[\frac{1}{\frac{1+x^4-2x^2+3x^2}{(1-x^2)^2}}\right]\left[\frac{\sqrt{3}+\sqrt{3}x^2}{(1-x^2)^2}\right]$
=> $\frac{dy}{dx}=\frac{-2x^2+2}{x^4+x^2+1}+\frac{2}{\sqrt{3}}\left[\frac{(1-x^2)^2}{1+x^4-2x^2+3x^2}\right]\left[\frac{\sqrt{3}(1+x^2)}{(1-x^2)^2}\right]$
=> $\frac{dy}{dx}=\frac{-2x^2+2}{x^4+x^2+1}+\frac{2}{\sqrt{3}}\left[\frac{\sqrt{3}(1+x^2)}{1+x^4+x^2}\right]$
=> $\frac{dy}{dx}=\frac{-2x^2+2}{x^4+x^2+1}+\frac{2(1+x^2)}{x^4+x^2+1}$
=> $\frac{dy}{dx}=\frac{-2x^2+2+2(1+x^2)}{x^4+x^2+1}$
=> $\frac{dy}{dx}=\frac{-2x^2+2+2+2x^2}{x^4+x^2+1}$
=> $\frac{dy}{dx}=\frac{4}{x^4+x^2+1}$
Question 53. If y = (sin x − cos x)^{sin x−cos x}, π/4 < x < 3π/4, find $\frac{dy}{dx}$ .
Solution:
We have,
=> y = (sin x − cos x)^{sin x−cos x}
On taking log of both the sides, we get,
=> log y = log (sin x − cos x)^{sin x−cos x}
=> log y = (sin x − cos x) log (sin x−cos x)
On differentiating both sides with respect to x, we get,
=> $\frac{1}{y}\frac{dy}{dx}=(sinx−cosx)(\frac{1}{sinx-cosx})(cosx+sinx)+log (sin x−cos x)(cosx+sinx)$
=> $\frac{1}{y}\frac{dy}{dx}$ = (1)(cosx + sinx) + (cosx + sinx)log (sin x − cos x)
=> $\frac{1}{y}\frac{dy}{dx}$ = cosx + sinx + (cosx + sinx)log (sin x − cos x)
=> $\frac{1}{y}\frac{dy}{dx}$ = (cosx + sinx)(1 + log (sin x − cos x))
=> $\frac{dy}{dx}$ = y(cosx + sinx)(1 + log (sin x − cos x))
=> $\frac{dy}{dx}$ = (sinx – cosx)^sinx-cosx(cosx + sinx)(1 + log (sin x − cos x))
Question 54. Find dy/dx of function xy = e^x-y.
Solution:
We have,
=> xy = e^x-y
On taking log of both the sides, we get,
=> log xy = log e^x-y
=> log x + log y = (x − y) log e
=> log x + log y = x − y
On differentiating both sides with respect to x, we get,
=> $\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}= 1 −\frac{dy}{dx}$
=> $\frac{1}{y}\frac{dy}{dx}+\frac{dy}{dx}= 1-\frac{1}{x}$
=> $(\frac{1}{y}+1)\frac{dy}{dx}= 1-\frac{1}{x}$
=> $(\frac{1+y}{y})\frac{dy}{dx}=\frac{x-1}{x}$
=> $\frac{dy}{dx}=\frac{y(x-1)}{x(y+1)}$
Question 55. Find dy/dx of function y^x + x^y + x^x = a^b.
Solution:
We have,
=> y^x + x^y+ x^x = a^b
=> $e^{logy^x}+e^{logx^y}+e^{logx^x} = a^b$
=> $e^{xlogy}+e^{ylogx}+e^{xlogx} = a^b$
On differentiating both sides with respect to x, we get,
=> $(e^{xlogy})[x(\frac{1}{y})(\frac{dy}{dx})+logy]+(e^{ylogx})[y(\frac{1}{x})+logx(\frac{dy}{dx})]+(e^{xlogx})[x(\frac{1}{x})+logx] = 0$
=> $(e^{xlogy})[\frac{x}{y}\frac{dy}{dx}+logy]+(e^{ylogx})[\frac{y}{x}+logx(\frac{dy}{dx})]+(e^{xlogx})[1+logx] = 0$
=> $(y^x)[\frac{x}{y}\frac{dy}{dx}+logy]+(y^x)[\frac{y}{x}+logx(\frac{dy}{dx})]+(x^x)[1+logx] = 0$
=> $(xy^{x-1}+x^ylogy)\frac{dy}{dx}=-x^x(1+logx)-yx^{y-1}-y^xlogy$
=> $(xy^{x-1}+x^ylogy)\frac{dy}{dx}=-[x^x(1+logx)+yx^{y-1}+y^xlogy]$
=> $\frac{dy}{dx}=-\frac{x^x(1+logx)+yx^{y-1}+y^xlogy}{xy^{x-1}+x^ylogy}$
Question 56. If (cos x)^y = (cos y)^x, find dy/dx.
Solution:
We have,
=> (cos x)^y = (cos y)^x
On taking log of both the sides, we get,
=> log (cos x)^y= log (cos y)^x
=> y log (cos x) = x log (cos y)
On differentiating both sides with respect to x, we get,
=> $y[(\frac{1}{cosx})(-sinx)]+log (cos x)(\frac{dy}{dx})=x[(\frac{1}{cosy})(-siny)(\frac{dy}{dx})]+log(cos y)$
=> $-ytanx+log (cos x)(\frac{dy}{dx})=-xtany(\frac{dy}{dx})+log(cos y)$
=> $log (cos x)(\frac{dy}{dx})+xtany(\frac{dy}{dx})=log(cos y)+ytanx$
=> $[log (cos x)+xtany]\frac{dy}{dx}=log(cos y)+ytanx$
=> $\frac{dy}{dx}=\frac{log(cos y)+ytanx}{log (cos x)+xtany}$
Question 57. If cos y = x cos (a+y), where cos a ≠ ±1, prove that $\frac{dy}{dx}=\frac{cos^2(a+y)}{sina}$ .
Solution:
We have,
=> cos y = x cos (a+y)
On differentiating both sides with respect to x, we get,
=> $(-siny)(\frac{dy}{dx})=x[-sin (a+y)(\frac{dy}{dx})]+cos(a+y)(1)$
=> $-siny(\frac{dy}{dx})=-xsin (a+y)(\frac{dy}{dx})+cos(a+y)$
=> $-siny\frac{dy}{dx}+xsin(a+y)\frac{dy}{dx}=cos(a+y)$
=> $[xsin(a+y)-siny]\frac{dy}{dx}=cos(a+y)$
=> $\frac{dy}{dx}=\frac{cos(a+y)}{xsin(a+y)-siny}$
=> $\frac{dy}{dx}=\frac{cos^2(a+y)}{cos(a+y)[xsin(a+y)-siny]}$
=> $\frac{dy}{dx}=\frac{cos^2(a+y)}{xcos(a+y)sin(a+y)-cos(a+y)siny}$
=> $\frac{dy}{dx}=\frac{cos^2(a+y)}{cosysin(a+y)-cos(a+y)siny}$
=> $\frac{dy}{dx}=\frac{cos^2(a+y)}{sin(a+y-y)}$
=> $\frac{dy}{dx}=\frac{cos^2(a+y)}{sina}$
Hence proved.
Question 58. If $(x-y)e^{\frac{x}{x-y}}=a$ , prove that $y\frac{dy}{dx}+x=2y$ .
Solution:
We have,
=> $(x-y)e^{\frac{x}{x-y}}=a$
On differentiating both sides with respect to x, we get,
=> $(x-y)\left[(e^{\frac{x}{x-y}})[\frac{(x-y)-x(1-\frac{dy}{dx})}{(x-y)^2}]\right]+e^{\frac{x}{x-y}}(1-\frac{dy}{dx})=0$
=> $\frac{(x-y)-x(1-\frac{dy}{dx})}{(x-y)}+(1-\frac{dy}{dx})=0$
=> $(1-\frac{dy}{dx})(1-\frac{x}{x-y})+1=0$
=> $(1-\frac{dy}{dx})(\frac{x-y-x}{x-y})+1=0$
=> $(1-\frac{dy}{dx})(\frac{-y}{x-y})+1=0$
=> $-y+y\frac{dy}{dx}+x-y=0$
=> $y\frac{dy}{dx}+x=2y$
Hence proved.
Question 59. If $x=e^{\frac{x}{y}}$ , prove that $\frac{dy}{dx}=\frac{x-y}{xlogx}$ .
Solution:
We have,
=> $x=e^{\frac{x}{y}}$
On taking log of both the sides, we get,
=> log x = log $e^{\frac{x}{y}}$
=> $logx=\frac{x}{y}loge$
=> $logx=\frac{x}{y}$
=> $y=\frac{x}{logx}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=\frac{logx(1)-x(\frac{1}{x})}{(logx)^2}$
=> $\frac{dy}{dx}=\frac{logx-1}{(logx)^2}$
We know, $y=\frac{x}{logx}$
=> $logx=\frac{x}{y}$
So, we get,
=> $\frac{dy}{dx}=\frac{\frac{x}{y}-1}{(logx)^2}$
=> $\frac{dy}{dx}=\frac{\frac{x-y}{y}}{(logx)^2}$
=> $\frac{dy}{dx}=\frac{x-y}{y(logx)^2}$
=> $\frac{dy}{dx}=\frac{x-y}{(\frac{x}{logx})(logx)^2}$
=> $\frac{dy}{dx}=\frac{x-y}{xlogx}$
Hence proved.
Question 60. If $y=x^{tanx}+\sqrt{\frac{x^2+1}{2}}$ , find dy/dx.
Solution:
We have,
=> $y=x^{tanx}+\sqrt{\frac{x^2+1}{2}}$
=> $y=e^{logx^{tanx}}+e^{log\sqrt{\frac{x^2+1}{2}}}$
=> $y=e^{tanxlogx}+e^{\frac{1}{2}log(\frac{x^2+1}{2})}$
On differentiating both sides with respect to x, we get,
=> $\frac{dy}{dx}=(e^{tanxlogx})[tanx(\frac{1}{x})+logx(sec^2x)]+(e^{\frac{1}{2}log(\frac{x^2+1}{2})})(\frac{1}{2})(\frac{2}{x^2+1})(\frac{1}{2})(2x)$
=> $\frac{dy}{dx}=(e^{tanxlogx})[\frac{tanx}{x}+sec^2xlogx]+(e^{\frac{1}{2}log(\frac{x^2+1}{2})})(\frac{x}{x^2+1})$
=> $\frac{dy}{dx}=x^{tanx}[\frac{tanx}{x}+sec^2xlogx]+\sqrt{\frac{x^2+1}{2}}(\frac{x}{x^2+1})$
=> $\frac{dy}{dx}=x^{tanx}[\frac{tanx}{x}+sec^2xlogx]+\frac{x}{\sqrt{2(x^2+1)}}$
Question 61. If $y=1+\frac{\alpha}{(\frac{1}{x}-\alpha)}+\frac{\frac{\beta}{x}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)}+\frac{\frac{\gamma}{x^2}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)(\frac{1}{x}-\gamma)}$ , find dy/dx.
Solution:
We are given,
=> $y=1+\frac{\alpha}{(\frac{1}{x}-\alpha)}+\frac{\frac{\beta}{x}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)}+\frac{\frac{\gamma}{x^2}}{(\frac{1}{x}-\alpha)(\frac{1}{x}-\beta)(\frac{1}{x}-\gamma)}$
Now we know,
If $y=1+\frac{ax^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{(x-c)}$ then, $\frac{dy}{dx}=\frac{y}{x}[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}]$
In the given expression, we have 1/x instead of x.
So, using the above theorem, we get,
=> $\frac{dy}{dx}=\frac{\alpha}{(\frac{1}{x}-\alpha)}+\frac{\beta}{(\frac{1}{x}-\beta)}+\frac{\gamma}{(\frac{1}{x}-\gamma)}$

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RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiatio

Question 1. Differentiate y = x1/x with respect to x.

Question 2. Differentiate y = xsin x with respect to x.

Question 3. Differentiate y = (1 + cos x)x with respect to x.

Question 4. Differentiate with respect to x.

Question 5. Differentiate y = (log x)x with respect to x.

Question 6. Differentiate y = (log x)cos x with respect to x.

Question 7. Differentiate y = (sin x)cos x with respect to x.

Question 8. Differentiate y = ex log x with respect to x.

Question 9. Differentiate y = (sin x)log x with respect to x.

Question 10. Differentiate y = 10log sin x with respect to x.

Question 11. Differentiate y = (log x)log x with respect to x.

Question 12. Differentiate with respect to x.

Question 13. Differentiate y = sin xx with respect to x.

Question 14. Differentiate y = (sin−1x)x with respect to x.

Question 15. Differentiate with respect to x.

Question 16. Differentiate with respect to x.

Question 17. Differentiate with respect to x.

Question 18. Differentiate the following with respect to x.

(i) y = xx √x

(ii)

(iii)

(iv) y = (x cos x)x + (x sin x)1/x

(v)

(vi) y = esin x + (tan x)x

(vii) y = (cos x)x + (sin x)1/x

(viii) , for x > 3

Question 19. Find dy/dx when y = ex + 10x + xx.

Question 20. Find dy/dx when y = xn + nx + xx + nn.

Question 21. Find dy/dx when .

Question 22. Find dy/dx when .

Question 23. Find dy/dx when y = e3x sin 4x 2x.

Question 24. Find dy/dx when y = sin x sin 2x sin 3x sin 4x.

Question 25. Find dy/dx when y = xsin x + (sin x)x.

Question 26. Find dy/dx when y = (sin x)cos x + (cos x)sin x.

Question 27. Find dy/dx when y = (tan x)cot x + (cot x)tan x.

Question 28. Find dy/dx when y = (sin x)x + sin−1 √x.

Question 29. Find dy/dx when

(i) y = xcos x + (sin x)tan x

(ii) y = xx + (sin x)x

Question 30. Find dy/dx when y = (tan x)log x + cos2 (π/4).

Question 31. Find dy/dx when .

Question 32. Find dy/dx when y = (log x)x+ xlogx.

Question 33. If x13y7 = (x+y)20, prove that .

Question 34. If x16y9 = (x2 + y)17, prove that .

Question 35. If y = sin xx, prove that .

Question 36. If xx + yx = 1, prove that .

Question 37. If xy × yx = 1, prove that .

Question 38. If xy + yx = (x+y)x+y, find dy/dx.

Question 39. If xm yn = 1, prove that .

Question 40. If yx = ey−x, prove that .

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Question 1. Differentiate y = x^1/x with respect to x.

Question 2. Differentiate y = x^{sin x} with respect to x.

Question 3. Differentiate y = (1 + cos x)^x with respect to x.

Question 4. Differentiate $y=x^{cos^{-1}x}$ with respect to x.

Question 5. Differentiate y = (log x)^x with respect to x.

Question 6. Differentiate y = (log x)^{cos x} with respect to x.

Question 7. Differentiate y = (sin x)^{cos x} with respect to x.

Question 8. Differentiate y = e^{x log x} with respect to x.

Question 9. Differentiate y = (sin x)^{log x} with respect to x.

Question 10. Differentiate y = 10^{log sin x} with respect to x.

Question 11. Differentiate y = (log x)^{log x} with respect to x.

Question 12. Differentiate $y = 10^{10^x}$ with respect to x.

Question 13. Differentiate y = sin x^x with respect to x.

Question 14. Differentiate y = (sin⁻¹x)^x with respect to x.

Question 15. Differentiate $y=x^{sin^{-1}x}$ with respect to x.

Question 16. Differentiate $y=(tanx)^{\frac{1}{x}}$ with respect to x.

Question 17. Differentiate $y=x^{tan^{-1}x}$ with respect to x.

(i) y = x^x √x

(ii) $y=x^{sinx-cosx}+\frac{x^2-1}{x^2+1}$

(iii) $y=x^{xcosx}+\frac{x^2+1}{x^2-1}$

(iv) y = (x cos x)^x + (x sin x)^1/x

(v) $y=(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}$

(vi) y = e^{sin x} + (tan x)^x

(vii) y = (cos x)^x + (sin x)^1/x

(viii) $y=x^{x^2-3}+(x-3)^{x^2}$ , for x > 3

Question 19. Find dy/dx when y = e^x + 10^x + x^x.

Question 20. Find dy/dx when y = xⁿ + n^x+ x^x + nⁿ.

Question 21. Find dy/dx when $y=\frac{(x^2-1)^3(2x-1)}{\sqrt{(x-3)(4x-1)}}$ .

Question 22. Find dy/dx when $y=\frac{e^{ax}secxlogx}{\sqrt{1-2x}}$ .

Question 23. Find dy/dx when y = e^3x sin 4x 2^x.

Question 25. Find dy/dx when y = x^{sin x} + (sin x)^x.

Question 26. Find dy/dx when y = (sin x)^{cos x} + (cos x)^{sin x}.

Question 27. Find dy/dx when y = (tan x)^{cot x} + (cot x)^{tan x}.

Question 28. Find dy/dx when y = (sin x)^x + sin⁻¹√x.

(i) y = x^{cos x} + (sin x)^{tan x}

(ii) y = x^x + (sin x)^x

Question 30. Find dy/dx when y = (tan x)^{log x} + cos² (π/4).

Question 31. Find dy/dx when $y=x^x+x^{\frac{1}{x}}$ .

Question 32. Find dy/dx when y = (log x)^x+ x^logx.

Question 33. If x¹³y⁷ = (x+y)²⁰, prove that $\frac{dy}{dx}=\frac{y}{x}$ .

Question 34. If x¹⁶y⁹ = (x² + y)¹⁷, prove that $x\frac{dy}{dx}=2y$ .

Question 35. If y = sin x^x, prove that $\frac{dy}{dx}=cos(x^x)×x^x(1+logx)$ .

Question 36. If x^x + y^x = 1, prove that $\frac{dy}{dx}=\frac{x^x(1+logx)+y^xlogy}{xy^{x-1}}$ .

Question 37. If x^y × y^x = 1, prove that $\frac{dy}{dx}=\frac{-y(y+xlogy)}{x(ylogx+x)}$ .

Question 38. If x^y+ y^x = (x+y)^x+y, find dy/dx.

Question 39. If x^m yⁿ = 1, prove that $\frac{dy}{dx}=-\frac{my}{nx}$ .

Question 40. If y^x = e^y−x, prove that $\frac{dy}{dx}=\frac{(1+logy)^2}{logy}$ .