Question

If $y=co{s}^{-1}\left(sin\right)$ then prove that $\frac{dy}{dx}=-1$

Answer 1

Given $y=co{s}^{-1}\left(sinx\right)$

on differentiating wrt 'x' we get

$\frac{dy}{dx}=\frac{d}{dx}\left[co{s}^{-1}\right(sinx\left)\right]$

$\frac{dy}{dx}=\frac{-1}{\sqrt{1-si{n}^2}}\times cosx$

$\Rightarrow \frac{dy}{dx}=\frac{-cosx}{\sqrt{co{s}^{2}x}}$ $[co{s}^{2}x+si{n}^{2}x=1]$

$\frac{dy}{dx}=\frac{-cosx}{cosx}=-1$

Hence $[\frac{dy}{dx}=-1]$