RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	11
Exercise	11.7
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation

Question 1. Find $\frac{dy}{dx}$ , when: x = at² and y = 2at

Solution:

Given that x = at², y = 2at

So,

$\frac{dx}{dt}=\frac{d}{dt}(at^2)=2at\\ \frac{dy}{dt}=\frac{d}{dt}(2at)=2a$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2a}{2at}=\frac{1}{t}$

Question 2. Find $\frac{dy}{dx}$ , when: x = a(θ + sinθ) and y = a(1 – cosθ)

Solution:

Here,

x = a(θ + sinθ)

Differentiating it with respect to θ,

$\frac{dx}{dθ}=a(1+cosθ)\ \ \ ..........(1)$

and,

y = a(1 – cosθ)

Differentiate it with respect to θ,

$\frac{dy}{dθ}=a(θ+sinθ)\\ \frac{dy}{dθ}=asinθ\ \ \ \ .....(2)$

Using equation (1) and (2),

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}\\ =\frac{asinθ}{a(1-cosθ)}\\ =\frac{\frac{2sinθ}{2}\frac{cosθ}{2}}{\frac{2sin^2θ}{2}},\ \ \ \ \ \ \ \left\{Since,\ 1-cosθ=\frac{2sin2θ}{2},\frac{2sinθ}{2}\frac{cosθ}{2}=sinθ\right\}\\ =\frac{dy}{dx}=\frac{tanθ}{2}$

Question 3. Find $\frac{dy}{dx}$ , when: x = acosθ and y = bsinθ

Solution:

Then x = acosθ and y = bsinθ

Then,

$\frac{dx}{dθ}=\frac{d}{dθ}(acosθ)=-asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(bsinθ)=bcosθ$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{bcosθ}{-asinθ}=-\frac{b}{a}cotθ$

Question 4. Find $\frac{dy}{dx}$ , when: x = ae^Θ(sinθ -cosθ), y = ae^Θ (sinθ +cosθ)

Solution:

Here,

x = ae^Θ (sinθ – cosθ)

Differentiating it with respect to θ,

$\frac{dx}{dθ}=a[e^θ\frac{d}{dθ}(sinθ-cosθ)+(sinθ-cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ+sinθ)+(sinθ-cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θsinθ]\ \ \ \ \ \ .......(1)$

And,

y = ae^Θ(sinθ+cosθ)

Differentiating it with respect to θ

$\frac{dy}{dθ}=a[e^θ\frac{d}{dθ}(sinθ+cosθ)+(sinθ+cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ-sinθ)+(sinθ+cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θcosθ]\ \ \ \ \ \ .......(2)$

Dividing equation (2) by equation (1)

$\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(2e^θ cosθ)}{a(2e^θ sinθ)}\\ \frac{dy}{dx}=cotθ$

Question 5. Find $\frac{dy}{dx}$ , when: x = bsin²θ and y = acos²θ

Solution:

Here,

x = bsin²θ and y = acos²θ

Then,

$\frac{d}{dθ}=\frac{d}{dθ}(bsin^2θ)=2bsinθ cosθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(acos^2θ)=-2acosθ sinθ \\ \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-2acosθ sinθ}{2bsinθ cosθ}=-\frac{a}{b}\\$

Question 6. Find $\frac{dy}{dx}$ , when: x = a(1 – cosθ) and y = a(θ +sinθ) at θ = $\frac{\pi}{2}$

Solution:

Here,

x = a(1 – cosθ) and y = a(θ + sinθ)

Then,

$\frac{dx}{dθ}=\frac{d}{dθ}[a(1-cosθ)]=asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}[a(θ +sinθ)]=a(1+cosθ)$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(1+cosθ)}{a(sinθ)}|_{θ =\frac{x}{2}}\\ =\frac{a(1+0)}{a}=1$

Question 7. Find $\frac{dy}{dx}$ , when: $x=\frac{e^t+e^{-t}}{2}$ and $y=\frac{e^t-e^{-t}}{2}$

Solution:

Here,

$x=\frac{e^t+e^{-t}}{2}$

Differentiate it with respect to t,

$\frac{dx}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)+\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t+e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t-e^{-t})=y\ \ \ ......(1)$

and,

$y=\frac{e^t-e^{-t}}{2}$

Differentiating it with respect to t,

$\frac{dy}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)-\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t-e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t+e^{-t})=x$

$\frac{dy}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)-\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t-e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t+e^{-t})=x\ \ \ \ .....(2)$

Dividing equation (2) and (1)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{x}{y}\\ \frac{dy}{dt}=\frac{x}{y}$

Question 8. Find $\frac{dy}{dx}$ , when: $x=\frac{3at}{1+t^2}$ and $y=\frac{3at^2}{1+t^2}$

Solution:

Here,

$x=\frac{3at}{1+t^2}$

Differentiating it with respect to t using quotient rule,

$\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at)-3at\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(3a)-3at(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{3a+3at^2-6at^2}{(1+t^2)^2}\right]\\ =\left[\frac{3a-3at^2}{(1-t^2)^2}\right]\\ \frac{dx}{dt}=\frac{3a(1-t^2)}{(1+t^2)^2}\ \ \ \ ....(1)$

and,

$y=\frac{3at^2}{1+t^2}$

Differentiating it with respect to t using quotient rule,

$\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at^2)-3at^2\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(6at)-(3at^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{6at+6at^3-6at^3}{(1+t^2)^2}\right]\\ \frac{dy}{dt}=\frac{6at}{(1+t^2)^2}\ \ \ \ ....(2)$

Dividing equation (2) by (1)

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6at}{(1+t^2)^2}\times\frac{(1+t^2)^2}{3a(1-t^2)}\\ \frac{dy}{dt}=\frac{2t}{1-t^2}$

Question 9. If x and y are connected parametrically by the equation, without eliminating the parameter, find $\frac{dy}{dx}$ when: x = a(cosθ +θsinθ), y = a(sinθ -θcosθ)

Solution:

The given equations are

x = a(cosθ +θ sinθ) and y = a(sinθ -θcosθ)

Then,

$\frac{dx}{dθ}=a\left[\frac{d}{dθ}cosθ +\frac{d}{dθ}(θ sinθ)\right]\\ =a\left[-sinθ +θ \frac{d}{dθ}(sinθ)+sinθ \frac{d}{dθ}(θ)\right]$

= a[-sinθ + θcosθ + sinθ] = aθcosθ

$\frac{dy}{dθ}=a\left[\frac{d}{dθ}sinθ +\frac{d}{dθ}(θ cosθ)\right]\\ =a\left[cosθ -\{θ \frac{d}{dθ}(cosθ)+cosθ \frac{d}{dθ}(θ)\}\right]$

= a[cosθ +θsinθ -cosθ]

= aθsinθ

Therefore,

$\frac{dy}{dx}=\frac{\left(\frac{dy}{dθ}\right)}{\left(\frac{dx}{dθ}\right)}=\frac{aθsinθ}{aθ cosθ}=tanθ$

Question 10. Find $\frac{dy}{dx}$ , when: $x=e^θ \left(θ +\frac{1}{θ}\right)$ and $y=e^{-θ} \left(θ -\frac{1}{θ}\right)$

Solution:

Here,

$x=e^θ \left(θ +\frac{1}{θ}\right)$

Differentiating it with respect to θ using product rule,

$\frac{dx}{dθ}=e^θ \frac{d}{dθ}\left(θ +\frac{1}{θ}\right)+\left(θ +\frac{1}{θ}\right)\frac{d}{dθ}(e^θ)\\ =e^θ \left(1-\frac{1}{θ^2}\right)+\left(\frac{θ ^2+1}{θ}\right)e^θ \\ =e^θ \left(\frac{θ ^2-1+θ ^3+θ}{θ ^2}\right)\\ \frac{dx}{dθ}=\frac{e^θ (θ ^3+θ ^2+θ -1)}{θ ^2}\ \ \ .....(1)$

and,

$y=e^θ \left(θ -\frac{1}{θ}\right)$

Differentiating it with respect to θ using product rule and chain rule

$\frac{dy}{dθ}=e^{-θ} \frac{d}{dθ}\left(θ -\frac{1}{θ}\right)+\left(θ -\frac{1}{θ}\right)\frac{d}{dθ}(e^{-θ})\\ =e^{-θ} \left(1+\frac{1}{θ^2}\right)+\left(θ -\frac{1}{θ}\right)e^{-θ} \\ =e^{-θ} \left(1+\frac{1}{θ ^2}-θ +\frac{1}{θ}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{θ ^2+1 -θ^3 +θ}{θ ^2}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{-θ ^3+θ ^2+θ +1}{θ ^2}\right)\ \ \ \ ......(2)$

Question 11. Find $\frac{dy}{dx}$ , when $x=\frac{2t}{1+t^2}$ and $y=\frac{1-t^2}{1+t^2}$
Solution:

Here,
$x=\frac{2t}{1+t^2}$
Differentiating it with respect to t using quotient rule,
$\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(2)-2t(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ =\left[\frac{2-2t^2}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2}\ \ \ \ \ ....(1)$
and,
$y=\frac{1-t^2}{1+t^2}$
Differentiating it with respect to t using quotient rule,
$\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(1-t^2)-(1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{-2t-2t^3-2t+2t^3}{(1+t^2)^2}\right]\\ \frac{dy}{dt}=\frac{-4t}{(1+t^2)^2}\ \ \ \ \ ....(2)$
Dividing equation (2) by (1)
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4t}{(1+t^2)^2}\times\frac{(1+t^2)^2}{2(1-t^2)}\\ =\frac{-2t}{1-t^2}\\ \frac{dy}{dx}=-\frac{x}{y}\ \ \ \ \left[Since,\ \frac{x}{y}=\frac{2t}{1+t^2}\times\frac{1+t^2}{1-t^2}=\frac{2t}{1-t^2}\right]$
Question 12. Find $\frac{dy}{dx}$ , when $x=cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ and $y=sin^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$
Solution:
Here,
$x=cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$
Differentiating it with respect to t using chain rule,
$\frac{dx}{dt}=\frac{-1}{\sqrt{1-\left(\frac{1}{1+t^2}\right)}^2}\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right)\\ =\frac{-1}{\sqrt{1-\frac{1}{(1+t^2)}}}\left[\frac{-1}{2(1+t^2)^{\frac{3}{2}}}\right]\frac{d}{dt}(1+t^2)\\ =\frac{(1+t^2)^{\frac{1}{2}}}{\sqrt{1+t^2-1}}\times\frac{-1}{2(1+t^2)^{\frac{3}{2}}}(2t)\\ =\frac{-t}{\sqrt{t^2}\times(1+t^2)}\\ \frac{dx}{dt}=\frac{-1}{1+t^2}\ \ \ \ \ ....(1)$
Now,
$y=sin^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$
Differentiating it with respect to t using chain rule,
$\frac{dy}{dt}=\frac{1}{\sqrt{1-\frac{1}{(\sqrt{1+t^2})^2}}}\times\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right)\ \ \ \ .....(2)$
Dividing equation (2) by (1)
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{(1+t^2)}\times\frac{(1+t^2)}{-1}\\ \frac{dy}{dx}=1$
Question 13. Find $\frac{dy}{dx}$ , when $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$
Solution:
Here,
$x=\frac{1-t^2}{1+t^2}$
Differentiating it with respect to t using quotient rule,
$\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(1-t^2)-(1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{-2t-2t^3-2t+2t^3}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{-4t}{(1+t^2)^2}\ \ \ \ \ ....(1)$
and,
$y=\frac{2t}{1+t^2}$
Differentiating it with respect to t using quotient rule,
$\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(2)-2t(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2}\ \ \ \ \ ....(2)$
Question 14. If x = 2cosθ – cos2θ and y = 2sinθ – sin2θ, prove that $\frac{dy}{dx}=tan\left(\frac{3\theta}{2}\right)$
Solution:
Here,
x = 2cosθ – cos2θ
Differentiating it with respect to θ using chain rule,
$\frac{dx}{dθ}=2(-sinθ )-(-sin2θ)\frac{d}{dθ} (2θ )\\ =-2sinθ +2sin2θ \\ \frac{dx}{dθ}=2(sin2θ -sinθ )\ \ \ \ \ ....(1)$
and,
y = 2sinθ – sin2θ
Differentiating it with respect to θ using chain rule,
$\frac{dy}{dθ}=2cosθ -cos2θ \frac{d}{dθ}(2θ )\\ =2cosθ -cos2θ (2)\\ =2cosθ -2cos2θ \\ \frac{dy}{dθ}=2(cosθ -cos2θ )\ \ \ \ \ ....(2)$
Dividing equation (2) by equation (1),
$\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{2(cosθ -cos2θ )}{2(sin2θ - sinθ )}\\ =\frac{cosθ -cos2θ}{sin2θ -sinθ}\\ \frac{dy}{dx}=\frac{-2sin\left(\frac{θ +2θ}{2}\right)sin\left(\frac{θ -2θ}{2}\right)}{2cos\left(\frac{2θ +θ}{2}\right)sin\left(\frac{2θ -θ}{2}\right)}\ \ \ \ \left[Since,\ sinA-sinB=2cos\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right),\ \ cosA-cosB=-2sin\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)\right]\\ =\frac{-sin\left(\frac{3θ )}{2}\right)\left(sin\left(\frac{-θ}{2}\right)\right)}{cos\left(\frac{3θ}{2}\right)sin\left(\frac{θ}{2}\right)}\\ =\frac{sin\left(\frac{3θ}{2}\right)}{cos\left(\frac{3θ}{2}\right)}\\ \frac{dy}{dx}=tan\left(\frac{3θ}{2}\right)$
Question 15. If x = e^cos2t and y = e^sin2t prove that, $\frac{dy}{dx}=\frac{y\ log\ x}{x\ log\ y}$
Solution:
Here,
x = e^cos2t
Differentiating it with respect to t using chain rule,
$\frac{dx}{dt}=\frac{d}{dt}(e^{cos2t})\\ =e^{cos2t}\frac{d}{dt}(cos2t)\\ =e^{cos2t}(-sin2t)\frac{d}{dt}(2t)\\ \frac{dx}{dt}=-sin2te^{cos2t}\ \ \ \ \ .....(1)$
and,
y = e^sin2t
Differentiating it with respect to t using chain rule,
$\frac{dy}{dt}=\frac{d}{dt}(e^{sin2t})\\ =e^{sin2t}\frac{d}{dt}(sin2t)\\ =e^{sin2t}(cos2t)\frac{d}{dt}(2t)\\ =e^{sin2t}(cos2t)(2)\\ \frac{dy}{dt}=2cos2te^{sin2t}\ \ \ \ \ .....(2)$
Dividing equation (2) by (1)
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2cos2te^{sin2t}}{-2sin2te^{cos2t}}\\ \frac{dy}{dx}=-\frac{ylogx}{xlogy}$
$\left[Since,\ x=e^{cos2t}\Rightarrow logx=cos2t\ \ y=e^{sin2t}\Rightarrow logy=sin2t\right]$
Question 16. If x = cos t and y = sin t, prove that $\frac{dy}{dx}=\frac{1}{\sqrt3}\ at\ t=\frac{2x}{3}$
Solution:
Here,
x = cos t
Differentiating it with respect to t,
$\frac{dx}{dt}=\frac{d}{dt}(cos\ t)\\ \frac{dx}{dt}=-sin\ t\ \ \ \ \ ....(1)$
and,
y = sin t
Differentiating it with respect to t,
$\frac{dx}{dt}=\frac{d}{dt}(sin\ t)\\ \frac{dx}{dt}=cos\ t\ \ \ \ \ ....(2)$
Dividing equation (2) by (1),
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{cos\ t}{-sin\ t}\\ \frac{dy}{dx}=-cot\ t\\ \left(\frac{dy}{dx}\right)=-cot\left(\frac{2\pi}{3}\right)\\ =-cot\left(\pi-\frac{\pi}{3}\right)\\ =-\left[-cot\left(\frac{\pi}{3}\right)\right]\\ =cot\left(\frac{\pi}{3}\right)\\ \frac{dy}{dx}=\frac{1}{\sqrt3}$
Question 17. If $x=a\left(t+\frac{1}{t}\right)$ and $y=a\left(t-\frac{1}{t}\right)$ , Prove that $\frac{dy}{dx}=\frac{x}{y}$
Solution:
Here,
$x=a\left(t+\frac{1}{t}\right)$
Differentiating it with respect to t,
$\frac{dx}{dt}=a\frac{d}{dt}\left(t+\frac{1}{t}\right)\\ =a\left(1-\frac{1}{t^2}\right)\\ \frac{dx}{dt}=a\left(\frac{t^2-1}{t^2}\right)\ \ \ \ \ .....(1)$
and,
$y=a\left(t-\frac{1}{t}\right)$
Differentiating it with respect to t,
$\frac{dy}{dt}=a\frac{d}{dt}\left(t-\frac{1}{t}\right)\\ =a\left(1+\frac{1}{t^2}\right)\\ \frac{dy}{dt}=a\left(\frac{t^2+1}{t^2}\right)\ \ \ \ \ .....(2)$
Dividing equation (2) by (1)
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=a\frac{(t^2+1)}{t^2}\times\frac{t^2}{a(t^2-1)}\\ \frac{dy}{dx}=\frac{t^2+1}{t^2-1}\\ \frac{dy}{dx}=\frac{x}{y}\ \ \ \ \ \ \left[Since,\ \frac{x}{y}=\frac{a(t^2+1)}{t}\times\frac{t}{a(t^2-1)}=\left(\frac{t^2+1}{t^2-1}\right)\right]$
Question 18. If $x=sin^{-1}\left(\frac{2t}{1+t^2}\right)$ and $y =tan^{-1}\left(\frac{2t}{1-t^2}\right)$ , -1 < 1 < 1, prove that $\frac{dy}{dx}=1$
Solution:
Here,
$x=sin^{-1}\left(\frac{2t}{1+t^2}\right)$
Put t = tan θ
$x=sin^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)\\ =sin^{-1}(sin2θ )\\ =2θ \ \ \ \ \ \left[Since,\ sin\ 2x=\frac{2tan\ x}{1+tan^2x}\right]\\ x=2(tan^{-1}t)\ \ \ \ \ [Since,\ t=sin\ θ ]$
Differentiating it with respect to t,
$\frac{dx}{dt}=\frac{2}{1+t^2}\ \ \ \ \ .....(1)$
Further,
$y =tan^{-1}\left(\frac{2t}{1-t^2}\right)$
Put t = tan θ
$y=tan^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)\\ =tan^{-1}(tan2θ )\\ =2θ \ \ \ \ \ \left[Since,\ tan\ 2x=\frac{2tan\ x}{1-tan^2x}\right]\\ y=2tan^{-1}t\ \ \ \ \ [Since,\ t=tan\ θ ]$
Differentiating it with respect to t,
$\frac{dy}{dt}=\frac{2}{1+t^2}\ \ \ \ \ \ ......(2)$
Dividing equation (2) by (1),
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{1+t^2}\times\frac{1+t^2}{2}\\ \frac{dy}{dx}=1$
Question 19. If x and y are connected parametrically by the equation, without eliminating the parameter, find $\frac{dy}{dx}$ , when: $x=\frac{sin^3t}{\sqrt{cos2t}}$ , $y=\frac{cos^3t}{\sqrt{cos2t}}$
Solution:
Here, the given equations are $x=\frac{sin^3t}{\sqrt{cos2t}}$ and $y=\frac{cos^3t}{\sqrt{cos2t}}$
Thus,
$\frac{dx}{dt}\frac{d}{dt}\left[\frac{sin^3t}{\sqrt{cos2t}}\right]\\ =\frac{\sqrt{cos2t}.\frac{d}{dt}(sin^3t)-sin^3t.\frac{d}{dt}\sqrt{cos2t}}{cos2t}\\ =\frac{\sqrt{cos2t}.3sin^2t.\frac{d}{dt}(sin\ t)-sin^3t\times\frac{1}{2\sqrt{cos\ 2t}}.\frac{d}{dt}(cos2t)}{cos2t}\\ =\frac{3\sqrt{cos\ 2t}.sin^2t\ cos\ t-\frac{sin^3t}{2\sqrt{cos\ 2t}}.(-2sin\ 2t)}{cos\ 2t}\\ =\frac{3cos\ 2t\ sin^2tcos\ t+sin^3tsin\ 2t}{cos\ 2t\sqrt{cos\ 2t}}\\ \frac{dy}{dt}=\frac{d}{dt}\left[\frac{cos^3t}{\sqrt{cos\ 2t}}\right]\\ =\frac{\sqrt{cos\ 2t}.\frac{d}{dt}(cos^3t)-cos^3t.\frac{d}{dt}(\sqrt{cos\ 2t})}{cos\ 2t}\\ =\frac{-3cos\ 2t.cos^2t.sin\ t+cos^3t\ sin\ 2t}{cos\ 2t.\sqrt{cos\ 2t}}$
Therefore,
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3cos\ 2t.cos^2t. sin\ t+cos^3t\ sin\ 2t}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t\ sin\ 2t}\\ =\frac{-3cos\ 2t\ cos^2t.sin\ t+cos^3t(2sin\ t\ cos\ t)}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t(2sin\ t+2sin^3t)}\\ =\frac{[-3(2cos^2t-1)cos\ t+2cos^3t]}{[3(1-2sin^2t)sin\ t+2sin^3t]}\ \ \ \ \ \ \ [cos\ 2t=(2cos^2t-1),\ cos\ 2t=(1-2sin^2t)]\\ =\frac{-4cos^3t+3cos\ t}{3sin\ t-4sin^3\ t}\\ =\frac{-cos\ 3t}{sin\ 3t}\ \ \ \ \ \ [cos\ 3t=4cos^3t3cos\ t,\ sin\ 3t=3sin\ t-4sin^3t]\\ =-cot3t$
Question 20. If $x=\left(t+\frac{1}{t}\right)^a$ and $y=a^{\left(t+\frac{1}{t}\right)}$ , find $\frac{dy}{dx}$
Solution:
Here,
$x=\left(t+\frac{1}{t}\right)^a$
Differentiating it with respect to t using chain rule,
$\frac{dx}{dt}=\frac{d}{dt}\left(t+\frac{1}{t}\right)^a\\ =a\left(t+\frac{1}{t}\right)^{a-1}\frac{d}{dt}\left(t+\frac{1}{t}\right)\\ \frac{dx}{dt}=a\left(t+\frac{1}{t}\right)^{1-1}\left(1-\frac{1}{t^2}\right)\ \ \ \ \ .....(1)$
And,
$y=a^{\left(t+\frac{1}{t}\right)}$
Differentiating it with respect to t using chain rule,
$\frac{dy}{dt}=\frac{d}{dt}a^{\left(t+\frac{1}{t}\right)}\\ =a^{\left(t+\frac{1}{t}\right)}\times log\ a\ \frac{d}{dt}\left(t+\frac{1}{t}\right)\\ \frac{dy}{dt}=a^{\left(t+\frac{1}{t}\right)}\times log\ a\left(1-\frac{1}{t^2}\right)\ \ \ \ \ ......(2)$
Dividing equation (2) by (1)
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a\left(1-\frac{1}{t^2}\right)}{a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)}\\ \frac{dy}{dx}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a}{a\left(t+\frac{1}{t}\right)^{a-1}}$

Question 21. If $x=a\left(\frac{1+t^2}{1-t^2}\right)$ and $y=\frac{2t}{1-t^2}$ , find $\frac{dy}{dx}$

Solution:

Here,

$x=a\left(\frac{1+t^2}{1-t^2}\right)$

Differentiate it with respect to t using chain rule,

$\frac{dx}{dt}=a\left[\frac{(1+t^2)\frac{d}{dt}(1+t^2)-(1+t^2)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(2t)-(1+t^2)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{2t-2t^2+2t+2t^3}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{4at}{(1-t^2)^2}\ \ \ \ \ ......(1)$

And,

$y=\frac{2t}{1-t^2}$

Differentiate it with respect to t using quotient rule,

$\frac{dy}{dt}=a\left[\frac{(1-t^2)\frac{d}{dt}(t)-(t)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(1)-(t)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{1-t^2+2t}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{2(1+t^2)}{(1-t^2)}\ \ \ \ \ ......(2)$

Question 22. Find $\frac{dy}{dx}$ , if y = 12(1 – cos t), x = 10(t – sin t), $-\frac{\pi}{2}<t<\frac{\pi}{2}$

Solution:

It is given that,

y = 12(1 – cos t),

x = 10(t – sin t)

Therefore,

$\frac{dx}{dt}=\frac{d}{dt}[10(t-sin\ t)]\\ =10.\frac{d}{dt}(t-sin\ t)\\ =10(1- cos\ t)$

$\frac{dy}{dt}=\frac{d}{dt}[12(t-cos\ t)]\\ =12.\frac{d}{dt}(1-cos\ t)\\ =12(0- (-sin\ t)\\ =12sin\ t$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{12sin\ t}{10(1-cos\ t)}\\ =\frac{12\times2sin\frac{t}{2}\times cos\frac{t}{2}}{10\times2\ sin^2\frac{t}{2}}\\ =\frac{6}{5}\ cot\ \frac{t}{2}$

Question 23. If x = a(θ – sin θ) and y = a(1 – cos θ), find $\frac{dy}{dx}$ , at θ = $\frac{\pi}{3}$

Solution:

Here,

x = a(θ – sin θ)

and

y = a(1 – cos θ)

Then,

$\frac{dx}{dθ}=\frac{d}{dθ}[a(θ-sin\ θ]\\ =a(1-cos\ θ)$

$\frac{dy}{dθ}=\frac{d}{dθ}[a(1+cos\ θ]\\ =a(-sin\ θ)$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-asin\ θ}{a(1-cos\ θ)}|_{θ=\frac{\pi}{3}}\\ =-\frac{sin\frac{\pi}{3}}{1-cos\frac{\pi}{3}}\\ =\frac{\frac{\sqrt3}{2}}{1-\frac{1}{2}}=-\sqrt3$

Question 24. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at t = $\frac{\pi}{4},\ \frac{dy}{dx}=\frac{b}{a}$

Solution:

Consider the given functions,

x = a sin 2t (1 + cos 2t)

and

y = b cos 2t (1 – cos 2t)

Write again the functions,

x = a sin 2t + $\frac{a}{2}$ sin 4t

Differentiate the above function with respect to t,

$\frac{dx}{dt}=2a\ cos\ 2t+2a\ cos\ 4t\ \ \ \ \ ....(1)\\$

y = b cos 2t (1 – cos 2t)

y = b cos 2t – b cos2 2t

$\frac{dy}{dx}=-2b\ sin\ 2t+2b\ cos\ 2t\ sin\ 2t\\=-2b\ sin\ 2t+b\ sin\ 4t\ \ \ \ ....(2)$

From equation (1) and (2)

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-2b\ sin\ 2t+b\ sin\ 4t}{2a\ cos\ 2t+2a\ cos\ 4t}\\ \therefore\frac{dy}{dx}|_{\frac{\pi}{4}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}|_{t=\frac{\pi}{4}}=\frac{-2b}{-2a}=\frac{b}{a}$

Question 25. If x = cos t (3 – 2cos²t) and y = sin t (3 – 2 sin²t), find the value of $\frac{dy}{dx}$ at t = $\frac{\pi}{4}$

Solution:

Here, the given function:

x = cos t (3 – 2cos2t)

x = cos t – 2cos3t

$\frac{dx}{dt}=-3sin\ t+6cos^2t\ \ \ \ ......(1)$

y = sin t (3 – 2 sin2t)

y = 3cos t – 2sin3t

$\frac{dy}{dt}=3cos\ t-6sin^2tcos\ t\ \ \ \ .....(2)\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{3cos\ t-6sin^tcos\ t}{-3sin\ t+6cos^2t\ sin\ t}\\ =\frac{3cos\ t(1-2sin^2t)}{3sin\ t(2cos^2t-1)}\\ =cot\ t\frac{(1-2(1-cos^2t))}{(2cos^2t-1)}\\ =cot\ t\\ \frac{dy}{dx}|_{\frac{\pi}{4}}=cot\frac{\pi}{4}=1$

Question 26. If $x=\frac{1+log\ t}{t^2}$ , $y=\frac{3+2log\ t}{t}$ find $\frac{dy}{dx}$

Solution:

Here,

$x=\frac{1+log\ t}{t^2}$

and

$y=\frac{3+2log\ t}{t}$

$\frac{dx}{dt}=\frac{t^2\left(\frac{1}{t}\right)-(1+log\ t)(2t)}{t^4}\\ =\frac{t-2t-2tlog\ t}{t^4}\\ =\frac{-2log\ t-1}{t^3}$

$\frac{dy}{dt}=\frac{t\left(\frac{2}{t}\right)-(-3+2log\ t)(1)}{t^2}\\ =\frac{2-3-2tlog\ t}{t^2}\\ =\frac{-2log\ t-1}{t^2}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-2log\ t-1}{t^2}}{\frac{-2log\ t-1}{t^3}}=t$

Question 27. If x = 3sin t – sin3t, y = 3cos t – cos3t, find $\frac{dy}{dx}\ at\ t=\frac{\pi}{3}$

Solution:

x = 3sin t – sin3t

and,

y = 3cos t – cos3t

$\frac{dx}{dt}=3cos\ t-3cos3t\\ \frac{dy}{dt}=-3sin\ t+3sin3t\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3sin\ t+3sin3t}{3cos\ t-3cos3t}$

When, $t=\frac{\pi}{3}$

$\frac{dy}{dx}=\frac{-3sin(\frac{\pi}{3})+3sin(\pi)}{3cos(\frac{\pi}{3})-3cos(\pi)}=\frac{-3\times\frac{\sqrt3}{2}+0}{3\times\frac{1}{2}-3(-1)}=-\frac{1}{\sqrt3}$

Question 28. If $sin\ x=\frac{2t}{1+t^2}$ , $tan\ y=\frac{2t}{1-t^2}$ find $\frac{dy}{dx}$

Solution:

$sin\ x=\frac{2t}{1+t^2}$

and,

$tan\ y=\frac{2t}{1-t^2}$

$\Rightarrow x=sin^{-1}\left(\frac{2t}{1+t^2}\right)$

and

$\Rightarrow y=tan^{-1}\left(\frac{2t}{1-t^2}\right)$

$\frac{dx}{dt}=\frac{1}{\sqrt{1-\left(\frac{2t}{1+t^2}\right)^2}}\times\frac{2(1+t^2)-(2t)(2t)}{(1+t^2)^2}\\ \frac{dx}{dt}=\frac{2}{1+t^2}\\ \frac{dy}{dt}=\frac{1}{\left(\frac{2t}{1-t^2}\right)2+1}\times\frac{2(1-t^2)-(2t)(-2t)}{(1-t^2)^2}\\ \frac{dy}{dt}=\frac{2}{1+t^2}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{2}{(1+t^2)}}{\frac{2}{(1+t^2)}}=1$

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RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation

Question 1. Find , when: x = at2 and y = 2at

Question 2. Find , when: x = a(θ + sinθ) and y = a(1 – cosθ)

Question 3. Find , when: x = acosθ and y = bsinθ

Question 4. Find , when: x = aeΘ (sinθ -cosθ), y = aeΘ (sinθ +cosθ)

Question 5. Find , when: x = bsin2θ and y = acos2θ

Question 6. Find , when: x = a(1 – cosθ) and y = a(θ +sinθ) at θ =

Question 7. Find , when: and

Question 8. Find , when: and

Question 9. If x and y are connected parametrically by the equation, without eliminating the parameter, find when: x = a(cosθ +θsinθ), y = a(sinθ -θcosθ)

Question 10. Find , when: and

Question 21. If and , find

Question 22. Find , if y = 12(1 – cos t), x = 10(t – sin t),

Question 23. If x = a(θ – sin θ) and y = a(1 – cos θ), find , at θ =

Question 24. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at t =

Question 25. If x = cos t (3 – 2cos2t) and y = sin t (3 – 2 sin2t), find the value of at t =

Question 26. If , find

Question 27. If x = 3sin t – sin3t, y = 3cos t – cos3t, find

Question 28. If , find

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Question 1. Find $\frac{dy}{dx}$ , when: x = at² and y = 2at

Question 2. Find $\frac{dy}{dx}$ , when: x = a(θ + sinθ) and y = a(1 – cosθ)

Question 3. Find $\frac{dy}{dx}$ , when: x = acosθ and y = bsinθ

Question 4. Find $\frac{dy}{dx}$ , when: x = ae^Θ(sinθ -cosθ), y = ae^Θ (sinθ +cosθ)

Question 5. Find $\frac{dy}{dx}$ , when: x = bsin²θ and y = acos²θ

Question 6. Find $\frac{dy}{dx}$ , when: x = a(1 – cosθ) and y = a(θ +sinθ) at θ = $\frac{\pi}{2}$

Question 7. Find $\frac{dy}{dx}$ , when: $x=\frac{e^t+e^{-t}}{2}$ and $y=\frac{e^t-e^{-t}}{2}$

Question 8. Find $\frac{dy}{dx}$ , when: $x=\frac{3at}{1+t^2}$ and $y=\frac{3at^2}{1+t^2}$

Question 9. If x and y are connected parametrically by the equation, without eliminating the parameter, find $\frac{dy}{dx}$ when: x = a(cosθ +θsinθ), y = a(sinθ -θcosθ)

Question 10. Find $\frac{dy}{dx}$ , when: $x=e^θ \left(θ +\frac{1}{θ}\right)$ and $y=e^{-θ} \left(θ -\frac{1}{θ}\right)$

Question 21. If $x=a\left(\frac{1+t^2}{1-t^2}\right)$ and $y=\frac{2t}{1-t^2}$ , find $\frac{dy}{dx}$

Question 22. Find $\frac{dy}{dx}$ , if y = 12(1 – cos t), x = 10(t – sin t), $-\frac{\pi}{2}<t<\frac{\pi}{2}$

Question 23. If x = a(θ – sin θ) and y = a(1 – cos θ), find $\frac{dy}{dx}$ , at θ = $\frac{\pi}{3}$

Question 24. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at t = $\frac{\pi}{4},\ \frac{dy}{dx}=\frac{b}{a}$

Question 25. If x = cos t (3 – 2cos²t) and y = sin t (3 – 2 sin²t), find the value of $\frac{dy}{dx}$ at t = $\frac{\pi}{4}$

Question 26. If $x=\frac{1+log\ t}{t^2}$ , $y=\frac{3+2log\ t}{t}$ find $\frac{dy}{dx}$

Question 27. If x = 3sin t – sin3t, y = 3cos t – cos3t, find $\frac{dy}{dx}\ at\ t=\frac{\pi}{3}$

Question 28. If $sin\ x=\frac{2t}{1+t^2}$ , $tan\ y=\frac{2t}{1-t^2}$ find $\frac{dy}{dx}$