If y= ^{-1}≤ft\{{{1+x²}-{1-x²}}{{1+x²}+{1-x²}}\}, where |x|

Correct Answer is : x1−x4{x}{{1-x^{4}}}1−x4x

Correct Answer is : x1−x4{x}{{1-x^{4}}}1−x4x

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MHTCET 13 Apr 2026 Shift 1 Paper

Question : 118 of 150

Marks: +1, -0

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

|x| < 1

\frac{d y}{d x}

is equal to

Solution:

Concept:

Use substitution and trigonometric identities to simplify the expression for

y

before differentiating.

Explanation:

Let

a = \sqrt{1+x^{2}}

and

b = \sqrt{1-x^{2}}

Then

y = \tan^{-1}\left(\frac{a-b}{a+b}\right)

Set

t = \tan y = \frac{a-b}{a+b}

Use the identity

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

Compute

1-t^{2} = \frac{(a+b)^{2} - (a-b)^{2}}{(a+b)^{2}} = \frac{4ab}{(a+b)^{2}}

Therefore,

\tan 2y = \frac{2(a-b)}{a+b} \cdot \frac{(a+b)^{2}}{4ab} = \frac{a^{2}-b^{2}}{2ab}

Now,

a^{2} - b^{2} = (1+x^{2}) - (1-x^{2}) = 2x^{2}

Also,

ab = \sqrt{1+x^{2}}\sqrt{1-x^{2}} = \sqrt{1-x^{4}}

So,

\tan 2y = \frac{2x^{2}}{2\sqrt{1-x^{4}}} = \frac{x^{2}}{\sqrt{1-x^{4}}}

Since

|x| < 1

, we have

x^{2} \in [0,1)

, so

2y = \sin^{-1}(x^{2})

Thus,

y = \frac{1}{2}\sin^{-1}(x^{2})

Differentiating with respect to

x

\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1-x^{4}}} \cdot 2x = \frac{x}{\sqrt{1-x^{4}}}

Answer:

\frac{dy}{dx} = \frac{x}{\sqrt{1-x^{4}}}

, hence Option B.

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