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Learn Exercise 5.3 with Free Lessons & Tips

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Question 1:

Find :

Solution 1:

Comments

Gayathri Seethamraju

Question 2:

Find :

Solution 2:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 3:

Find :

Solution 3:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 4:

Find :

Solution 4:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain and

From (1) and (2), we obtain

Comments

Swapna Shree

Question 5:

Find :

Solution 5:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 6:

Find :

Solution 6:

The given relationship is

Differentiating this relationship with respect to x, we obtain

[Derivative of constant function is 0]

Comments

Swapna Shree

Question 7:

Find :

Solution 7:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 8:

Find :

Solution 8:

Step-by-step explanation:

sin²y + cos xy = pi

differentiating both side w.r.t.x.

$\frac{d(sin ^{2}+cos xy)}{dx}=\frac{d(k)}{dx}$

hence derivative of constant is zero

$\frac{d(sin^{2}y)}{dx} +\frac{d(cos xy)}{dx}=0$

calculating derivative of sin²y and cos(xy) sperately

calculating derivative of sin²y

$\frac{d(sin^2 y)}{dx}= \frac{d(sin^{2}y)}{dx} *\frac{dy}{dy}$

$= \frac{d(sin^{2}y)}{dx} *\frac{dy}{dx}$

$=2sin^{2-1}x.\frac{d(sin y)}{dy} * \frac{dy}{dx}$

= $=2siny.cosy*\frac{dy}{dx}$

calculating derivative of cos(xy)

$\frac{d(cos(xy))}{dx}=-sin(xy)*\frac{d}{dx} (xy)$

$=-sin(xy)*(\frac{d(x)}{dx}.y+\frac{d(y)}{d(x)}.x )$

$=-sin(xy).(1.y+x\frac{dy}{dx} )$

$=-sin(xy)(y+x\frac{dy}{dx} )$

$=-sin(xy).y-sin(xy).x\frac{dy}{dx}$

$=-ysin(xy)-sin(xy).x\frac{dy}{dx}$

now,

$\frac{d(sin^{2}y)}{dx}+\frac{d(cos(xy))}{dx}=0$

putting values

$2siny.cosy.\frac{dy}{dx} +(-ysin(xy)-xsin(xy)\frac{dy}{dx} )=0$

$2siny.cosy.\frac{dy}{dx}-ysin(xy)-xsin(xy)\frac{dy}{dx}=0$

$2siny cosy\frac{dy}{dx}-xsin(xy)\frac{dy}{dx}=y sin(xy)$

$\frac{dy}{dx}(2sinycosy-xsin(xy))=ysin(xy)\$

$\frac{dy}{dx}=\frac{ysin(xy)}{2sinycosy-xsinxy}$

$\frac{dy}{dx}=\frac{ysin(xy)}{sin2y-xsinxy}$

Comments

Manisha S.

Question 9:

Find :

Solution 9:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 10:

Find :

Solution 10:

We have :

$y=sin^{-1}[\frac{2x}{1+x^{2}}]$

put $x=tan\theta \Rightarrow \theta =tan^{-1}x$

Now,

$y=sin^{-1}[\frac{2tan\theta }{1+tan^{2}\theta }]$

$\Rightarrow y=sin^{-1}(sin2\theta ),$ as $(sin2\theta =\frac{2tan\theta }{1+tan^{2}\theta })$

$\Rightarrow y=2\theta$ , as ( $sin^{-1}(sinx)=x$ )

$\Rightarrow y=2tan^{-1}x$

$\Rightarrow \frac{dy}{dx}=2\times \frac{1}{1+x^{2}},$ , {because $\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^{2}}$ }

$\Rightarrow \frac{dy}{dx}=\frac{2}{1+x^{2}}$

Comments

Abhay

Question 11:

Find :

Solution 11:

The given relationship is

It is known that,

Comparing equations (1) and (2), we obtain

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 12:

Find :

Solution 12:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Using chain rule, we obtain

From (1), (2), and (3), we obtain

Alternate method

⇒

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 13:

Find :

Solution 13:

The given relationship is

Differentiating this relationship with respect to x, we obtain

Comments

Swapna Shree

Question 14:

Find :

Solution 14:

Steps to follow :

1st step put x= $cos\theta$

Then the denominator of bracket become the formula for cos2( $\theta$ )

Step 2 :invese of cos2( $\theta$ ) becomes sec2( $\theta$ )

Step 3:y=2( $\theta$ )

Step4 :put the value of $\theta$ in term of x

Step 5:now you must get something like y=2cos`(x)

Step 6: Differentiating this relationship with respect to x, we obtain

Comments

Somesh Shinde

All Exercises - Chapter 5 - Continuity and Differentiability

Exercise 5.3

Miscellaneous Exercise 5

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Learn Exercise 5.3 with Free Lessons & Tips

All Exercises - Chapter 5 - Continuity and Differentiability

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