If y= ^{-1}≤ft({25-x²}{25+x²}), then y^{}(1) is...

Correct Answer is : −513-5/13−135

Correct Answer is : −513-5/13−135

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MHTCET 19 Apr 2026 Shift 2 Paper

Question : 122 of 150

Marks: +1, -0

y=\sin ^{-1}\left(\frac{25-x^{2}}{25+x^{2}}\right)

y^{\prime}(1)

is...

Solution:

Concept:

Differentiate an inverse trigonometric function using the chain rule.

Explanation:

Let

u = \dfrac{25-x^2}{25+x^2}

, so

y = \sin^{-1}u

Then

y' = \dfrac{u'}{\sqrt{1-u^2}}

Using the quotient rule,

u' = \dfrac{(-2x)(25+x^2) - (25-x^2)(2x)}{(25+x^2)^2}

This simplifies to

u' = \dfrac{-100x}{(25+x^2)^2}

Now compute the denominator:

\sqrt{1-u^2} = \sqrt{1 - \left(\dfrac{25-x^2}{25+x^2}\right)^2} = \dfrac{10|x|}{25+x^2}

x=1

u' = \dfrac{-100}{26^2}

and

\sqrt{1-u^2} = \dfrac{10}{26}

Therefore,

y'(1) = \dfrac{-100/26^2}{10/26} = -\dfrac{5}{13}

Answer:

-\dfrac{5}{13}

(Option A).

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