Question

If $y={\cos }^{-1}\left[\frac{3x+4\sqrt{1-x^2}}{5}\right]$. Find $\frac{dy}{dx}$.

Answer 1

Now,

$y={\cos }^{-1}\left[\frac{3x+4\sqrt{1-x^2}}{5}\right]$

or, $y={\cos }^{-1}\left[\frac{3x}{5}+\sqrt{1-{\left(\frac{3}{5}\right)}^2}\sqrt{1-x^2}\right]$

or, $y={\cos }^{-1}\frac{3}{5}-{\cos }^{-1}x$

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$.