The derivative of ^{-1}≤ft({{1+x²}-1}{x}) with respect to ^{-1}≤ft({x}{{1-x²}}) at x=1/2 is

Correct Answer is : 35{{3}}{5}53

Correct Answer is : 35{{3}}{5}53

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MHTCET 17 Apr 2026 Shift 2 Paper

Question : 121 of 150

Marks: +1, -0

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

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

x=\frac{1}{2}

Solution:

Concept:

Use

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

to differentiate the given functions with respect to each other.

Explanation:

Let

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

and

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

Rationalise the expression inside

y

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

Using the identity

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

, we get

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

Therefore,

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

Now simplify

z

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

So,

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

Hence,

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

x = \frac{1}{2}

\frac{dy}{dz} = \frac{\sqrt{1-\frac{1}{4}}}{2\left(1+\frac{1}{4}\right)} = \frac{\frac{\sqrt{3}}{2}}{\frac{5}{2}} = \frac{\sqrt{3}}{5}

Answer:

\frac{\sqrt{3}}{5}

, which is Option C.

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