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Question

Mathematics Question on Continuity and differentiability

If y=cot⁡−1 (x2)y = \cot^{-1} \, (x^2), then the value of dydx \frac{dy}{dx} is equal to

A

2x1+x4\frac{2x}{1 + x^4}

B

2x1+4x\frac{2x}{\sqrt{1 + 4x}}

C

−2x1+x4\frac{-2x}{1 + x^4}

D

−2x1+x2\frac{- 2x}{\sqrt{1 + x^2}}

Answer

−2x1+x4\frac{-2x}{1 + x^4}

Explanation

Solution

Let y=cot⁡−1(x2)y =\cot^{-1} (x^2) ⇒   cot⁡ y=x2\Rightarrow \:\:\: \cot \, y = x^2 Diff both side, w.r.t. ′x′−cosec2y.dydx=2x'x' - cosec^2 y. \frac{dy}{dx} = 2x dydx=2x−cosec2y\frac{dy}{dx} = \frac{2x}{-cosec^2 y} =2x−(1+cot⁡2y)=2x−(x4+1)=−2x(x4+1) =\frac{2x}{-(1+ \cot^2 y)} = \frac{2x}{-(x^4 + 1)} = \frac{-2x}{(x^4 + 1)}