Find $\frac{dy}{dx}$ where $y=\sin^{-1}(

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

Find $\frac{dy}{dx}$ where $y=\sin^{-1}(\frac{1-x^2}{1+x^2})$

Options:

$\frac{1}{1+x}$

$x(1+x^2)$

$\frac{-2}{1+x^2}$

$1/x$

Correct Answer:

$\frac{-2}{1+x^2}$

Explanation:

$y=\sin^{-1}(\frac{1-x^2}{1+x^2})=\frac{π}{2}-\cos^{-1}(\frac{1-x^2}{1+x^2})$

$y=\frac{π}{2}-2\tan^{-1}x$

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