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Question

Mathematics Question on Derivatives

If y=sin⁡2cot⁡−11+x1−x,y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+x}{1-x}}, then dydx\frac{dy}{dx} is equal to

A

2sin⁡2x2\sin 2x

B

sin⁡2x\sin 2x

C

12\frac{1}{2}

D

−12-\frac{1}{2}

Answer

−12-\frac{1}{2}

Explanation

Solution

y=sin⁡2cot⁡−11+x1−xy={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+x}{1-x}} Put, x=cos⁡θx=\cos \theta
⇒\Rightarrow θ=cos⁡−1x\theta ={{\cos }^{-1}}x ...(i) y=sin⁡2cot⁡−11+cos⁡θ1−cos⁡θy={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }} y=sin⁡2cot⁡−12cos⁡2θ/22sin⁡2θ/2y={{\sin }^{2}}{{\cot }^{-1}}\sqrt{\frac{2{{\cos }^{2}}\theta /2}{2{{\sin }^{2}}\theta /2}} y=sin⁡2cot⁡−1(cot⁡θ/2)y={{\sin }^{2}}{{\cot }^{-1}}(\cot \theta /2) y=sin⁡2θ/2y={{\sin }^{2}}\theta /2 y=1−cos⁡θ2y=\frac{1-\cos \theta }{2} y=1−x2y=\frac{1-x}{2} [From E(i)] Differentiating w.r.t. x,x, we get dydx=−12\frac{dy}{dx}=-\frac{1}{2}