Question

If y = sin $(si{n}^{-1}x+co{s}^{-1}x),x\in [-1,1]then\frac{dy}{dx}$ is

• A. $\frac{\pi }{2}$
• B. $\frac{-\pi }{2}$
• C. 0
• D. 1

Answer 1

Answer: (C)

0

$y=\sin ({\sin }^{-1}x+{\cos }^{-1}x)\left(\text{Given}\right)$

Differentiating above equation w.r.t. $x$, we have

$\frac{dy}{dx}=\cos ({\sin }^{-1}x+{\cos }^{-1}x)[\frac{1}{\sqrt{1-x^2}}+\left(\frac{-1}{\sqrt{1-x^2}}\right)]$

$\Rightarrow \frac{dy}{dx}=\cos ({\sin }^{-1}x+{\cos }^{-1}x)[\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}]$

$\Rightarrow \frac{dy}{dx}=0$

Hence the value of $\frac{dy}{dx}$ for the given function is $0$.