# NCERT Solutions for Exercise 5.6 Class 12 Maths Chapter 5 - Continuity and Differentiability

In the previous exercises of this NCERT syllabus Class 12 chapter, you have learned about the differentiation of function y = f(x) w.r.t the x, but in exercise 5.6 Class 12 Maths, the function y = f(a) and x = f(a) is given in the parametric form and you need to find the differentiation of function y w.r.t the x. If you have solved the previous exercises by yourself, you won't find any difficulty while solving Class 12th Maths chapter 5 exercise 5.6. You just need to find the differentiation of y and x w.r.t. to its parameter. The differentiation of y w.r.t. x is the same as the ratio of dy/da and dx/da i.e dy/dx = (dy/da)/(dx/da). You can go through the NCERT solutions for Class 12 Maths chapter 5 exercise 5.6 to get detailed solutions to these problems. Click on the link to get NCERT Solutions from Class 6 to Class 12 at one place.

Also, see

• Continuity and Differentiability Exercise 5.1
• Continuity and Differentiability Exercise 5.2
• Continuity and Differentiability Exercise 5.3
• Continuity and Differentiability Exercise 5.4
• Continuity and Differentiability Exercise 5.5
• Continuity and Differentiability Exercise 5.7
• Continuity and Differentiability Exercise 5.8
• Continuity and Differentiability Miscellaneous Exercise

## Continuity and Differentiability Exercise: 5.6

Question:1 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

$x = 2at^2, y = at^4$

Given equations are
$x = 2at^2, y = at^4$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(2at^2)}{dt}= 4at$
Similarly,
$\frac{dy}{dt}=\frac{d(at^4)}{dt}= 4at^3$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{4at^3}{4at} = t^2$
Therefore, the answer is $\frac{dy}{dx}= t^2$

### Question:2 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

$x= a \cos \theta , y = b \cos \theta$

Given equations are
$x= a \cos \theta , y = b \cos \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\cos \theta)}{d\theta}= -a\sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\cos \theta)}{d\theta}= -b\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-b\sin \theta}{-a\sin \theta} = \frac{b}{a}$
Therefore, answer is $\frac{dy}{dx}= \frac{b}{a}$

### Question:3 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx . $x = \sin t , y = \cos 2 t$

Given equations are
$x = \sin t , y = \cos 2 t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t$
Similarly,
$\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t \ \ \ \ \ (\because \sin 2x = \sin x\cos x)$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t$
Therefore, the answer is $\frac{dy}{dx} = -4\sin t$

### Question:4 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx

$x = 4t , y = 4/t$

Given equations are
$x = 4t , y = 4/t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(4 t)}{dt}= 4$
Similarly,
$\frac{dy}{dt}=\frac{d(\frac{4}{t})}{dt}= \frac{-4}{t^2}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{ \frac{-4}{t^2} }{4} = \frac{-1}{t^2}$
Therefore, the answer is $\frac{dy}{dx} = \frac{-1}{t^2}$

### Question:5 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta$

Given equations are
$x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$
Therefore, answer is $\frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}$

### Question:6 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )$

Given equations are
$x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(1+\cos \theta))}{d\theta}=-a\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-a\sin \theta}{a(1-\cos \theta)} = \frac{-\sin }{1-\cos \theta} =- \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})$
Therefore, the answer is $\frac{dy}{dx}=-\cot \frac{\theta}{2}$

## Question:7 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$

Given equations are
$x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(\frac{\sin ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\sin^3t)}{dt}-\sin^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\sin^2 t\cos t.\sqrt{\cos 2t}-\sin^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{{\cos 2t}}$
$=\frac{3\sin^2t\cos t . \cos 2t+sin^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}} \ \ \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)$
Similarly,
$\frac{dy}{dt}=\frac{d( \frac{\cos ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\cos^3t)}{dt}-\cos^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\cos^2 t(-\sin t).\sqrt{\cos 2t}-\cos^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{(\sqrt{\cos 2t})^2}$
$=\frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin 2t}{\cos2t\sqrt{\cos2t}}$
$=\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}} }{\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}}} = \frac{\cot^3t(1-3\tan t \cot 2t)}{(3\cot t \cot 2t+1)}$
$= \frac{\cos^3t(1-3.\frac{\sin t}{\cos t}.\frac{\cos2t}{\sin 2t})}{\sin^3t(3.\frac{\cos t}{\sin t}.\frac{\cos 2t}{\sin 2t}+1)} = \frac{\cos^2t(\cos t\sin2t -3\sin t \cos 2t)}{\sin^2t(3\cos t \cos2t+\sin t \sin 2t)}$
$=\frac{\cos^2t(\cos t .2\sin t \cos t - 3\sin t (2\cos^2t-1))}{\sin^2t(3\cos t(1-2\sin^2 2t)+\sin t.2\sin t \cos t)}$
$(\because \sin 2x = 2\sin x\cos x \ and \ \cos 2x = 2\cos^2x-1 \ and \ \cos 2x = 1-2\sin^2x)$
$=\frac{\cos^2t(2\sin t\cos^2 t-6\sin t\cos^2t+3\sin t)}{\sin^2t(3\cos t-6\cos t \sin^2t+2\sin^2\cos t)}\\=\frac{sint cost(-4cos^3t+3cost)}{sintcost(3sint-4sin^3t)}$

$\frac{dy}{dx} = \frac{-4\cos^3t+3\cos t}{3\sin t -4\sin^3 t}= \frac{-\cos 3t}{\sin 3t} = -\cot 3t$ $\left ( \because \sin3t = 3\sin t-4\sin^3t \\ \ and \ \cos3t = 4\cos^3t - 3\cos t \right )$

Therefore, the answer is $\frac{dy}{dx} = -\cot 3t$

## Question:8 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = a ( \cos t + \log \tan t/2 ),y = a \sin t$

Given equations are
$x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t$
Now, differentiate both w.r.t t
We get,
$\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})$
$= a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})$
$=a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})$
Similarly,
$\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t$
Therefore, the answer is $\frac{dy}{dx} = \tan t$

## Question:9 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = a \sec \theta , y = b \ tan \theta$

Given equations are
$x = a \sec \theta , y = b \ tan \theta$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}$
Therefore, the answer is $\frac{dy}{dx} = \frac{b cosec \theta}{a}$

### Question:10 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx $x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )$

Given equations are
$x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )$
Now, differentiate both w.r.t $\theta$
We get,
$\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta$
Similarly,
$\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta$
Therefore, the answer is $\frac{dy}{dx}= \tan \theta$

### Question:11 If $x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$, show that $dy/dx = - y /x$

Given equations are
$x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}$

$xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\ sin^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c$

differentiating with respect to x

$x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x}$

## More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6:-

In Class 12th Maths chapter 5 exercise 5.6, there are 11 questions in which you need to find the differentiation of y w.r.t x where x and y are given in the parametric form and you can't use the elimination of parameter method to these differentiation. The three examples given in the NCERT book prior to Class 12 Maths ch 5 ex 5.6 are related to differentiation in parametric form also. You can solve these examples to get more clarity.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6:-

• NCERT solutions for Class 12 Maths Chapter 5 exercise 5.6 are beneficial not only for this exercise but also to learn differentiation of functions in parametric function.
• Class 12 Maths chapter 5 exercise 5.6 solutions can be used as the revision notes

• Class 12 Maths ch 5 ex 5.6 solutions are designed in a descriptive manner so you use these solution to understand the concept easily.

Also see-

• NCERT Solutions for Class 12 Maths Chapter 5

• NCERT Exemplar Solutions Class 12 Maths Chapter 5

## NCERT Solutions of Class 12 Subject Wise

• NCERT Solutions for Class 12 Maths

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Chemistry

• NCERT solutions for class 12 Biology

## Subject Wise NCERT Exampler Solutions

• NCERT Exemplar Solutions for Class 12th Maths

• NCERT Exemplar Solutions for Class 12th Physics

• NCERT Exemplar Solutions for Class 12th Chemistry

• NCERT Exemplar Solutions for Class 12th Biology

Happy learning!!!

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