RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiatio

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter11
Exercise11.5
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 11.5 Solutions Chapter 11 Differentiatio

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

Solution:

We have, 

=> y = x1/x

On taking log of both the sides, we get,

=> log y = log x1/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 = xsin x with respect to x.

Solution:

We have, 

=> y = xsin x

On taking log of both the sides, we get,

=> log y = log xsin 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 = ex log x with respect to x.

Solution:

We have, 

=> y=ex log x

=> y = e^{logx^x}

=> y = xx

On taking log of both the sides, we get,

=> log y = log xx

=> 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 = 10log sin x with respect to x.

Solution:

We have, 

=> y = 10log sin x

On taking log of both the sides, we get,

=> log y = log 10log 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 = 10x 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 xx with respect to x.

Solution:

We have, 

=> y = sin xx 

=> sin−1 y = xx

On taking log of both the sides, we get,

=> log (sin−1 y) = log xx

=> 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 = xx √x

Solution:

We have,

=> y = xx √x

On taking log of both the sides, we get,

=> log y = log (xx √x)

=> log y = log xx + 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 = esin x + (tan x)x

Solution:

We have, 

=> y = esin 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 = ex + 10x + xx.

Solution:

We have, 

=> y = ex + 10x + xx

=> 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+ xx + nn.

Solution:

We have, 

=> y = xn + nx + xx + 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 = e3x sin 4x 2x.

Solution:

We have 

=> y = e3x sin 4x 2x.

On taking log of both the sides, we get,

=> log y = log (e3x sin 4x 2x)

=> log y = log e3x + log (sin 4x) + log 2x

=> 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 = xsin x + (sin x)x.

Solution:

We have, 

=> y = xsin x + (sin x)x

Let u = xsin x and v = (sin x)x. Therefore, y = u + v.

Now, u = xsin x

On taking log of both the sides, we get,

=> log u = log xsin 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 = xcos x + (sin x)tan x

Solution:

We have, 

=> y = xcos 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 = xx + (sin x)x

Solution:

We have, 

=> y = xx + (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+ xlogx.

Solution:

We have, 

=> y = (log x)x+ xlogx

Let u = (log x)x and v = xlogx. 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 = xlogx

On taking log of both the sides, we get,

=> log v = log xlogx

=> 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 x13y= log (x+y)20

=> log x13 + log y= 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 = (x+ 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 xx, prove that \frac{dy}{dx}=cos(x^x)×x^x(1+logx)     .

Solution:

We have, 

=> y = sin xx 

Let u = xx. Now y = sin u.

On taking log of both the sides, we get,

=> log u = log xx

=> 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 xx + yx = 1, prove that \frac{dy}{dx}=\frac{x^x(1+logx)+y^xlogy}{xy^{x-1}}     .

Solution:

We have, 

=> xx + yx = 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 xy × yx = 1, prove that \frac{dy}{dx}=\frac{-y(y+xlogy)}{x(ylogx+x)}     .

Solution:

We have, 

=> xy × yx = 1

On taking log of both the sides, we get,

=> log (x× yx) = log 1

=> log x+ log yx = 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+ yx = (x+y)x+y, find dy/dx.

Solution:

We have, 

=> xy + yx = (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 xm yn = 1, prove that \frac{dy}{dx}=-\frac{my}{nx}     .

Solution:

We have, 

=> xm yn = 1

On taking log of both the sides, we get,

=> log (xm yn)= log 1

=> log xm + 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 yx = ey−x, prove that \frac{dy}{dx}=\frac{(1+logy)^2}{logy}     .

Solution:

We have, 

=> yx = ey−x

On taking log of both the sides, we get,

=> log yx = log ey−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 ex + ey = ex+y, prove that \frac{dy}{dx}+e^{y-x}=0      .
Solution:
We have,
=> ex + ey = ex+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 ey = yx, prove that \frac{dy}{dx}=\frac{(logy)^2}{logy-1}      .
Solution:
We have,
=> ey = yx
On taking log of both the sides, we get,
=> log ey = log yx
=> 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 ex+y − x = 0, prove that \frac{dy}{dx}=\frac{1-x}{x}      .
Solution:
We have,
=> ex+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 ex+y − x = 0
=> ex+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)
=> f'(x)=(1+x)(1+x^2)(1+x^4)(8x^7)+(1+x)(1+x^2)(1+x^8)(4x^3)+(1+x)(1+x^4)(1+x^8)(2x)+(1+x^2)(1+x^4)(1+x^8)(1)
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 = ex-y.
Solution:
We have,
=> xy = ex-y
On taking log of both the sides, we get,
=> log xy = log ex-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 yx + xy + xx = ab.
Solution:
We have,
=> yx + x+ xx = ab
=> 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)= 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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