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Manipal Entrance Test 2015 Math Paper

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Question : 21 of 70

Marks: +1, -0

The derivative of

\tan^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} \right)

with respect to

\tan^{-1} \left( \frac{2x\sqrt{1-x^{2}}}{1-2x^{2}} \right)

x=0,

is

$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{2}$
$1$

Solution:

Let

y=\tan^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} \right)

Putting x = tan θ, we get

y=\tan^{-1} \left( \frac{\sec\theta-1}{\tan\theta} \right)

= \tan^{-1} \left( \tan \frac{\theta}{2} \right)

= \frac{1}{2}\tan^{-1}x

On differentiating both sides w.r.t. x, we get

\frac{dy}{dx} = \frac{1}{2(1+x^{2})}

and

z=\tan^{-1} \left( \frac{2x\sqrt{1-x^{2}}}{1-2x^{2}} \right)

Putting

x=\sin\theta,

we get

z=\tan^{-1} \left( \frac{2\sin\theta\cos\theta}{\cos 2\theta} \right)

= \tan^{-1}(\tan 2\theta)

\Rightarrow z=2\theta=2\sin^{-1} x

On differentiating both sides w.r.t. x, we get

\frac{dz}{dx} = \frac{2}{\sqrt{1-x^{2}}}

Thus,

\frac{dy}{dz} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}}

= \frac{1}{4(1+x^{2})}\sqrt{1-x^{2}}

\therefore \left[ \frac{dy}{dz} \right]_{x=0} = \frac{1}{4}

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