Question

$\frac{d}{dx}\left({\cos }^{-1}(4x^3-3x)\right),$=$\therefore \frac{dy}{dx}=\frac{m}{\sqrt{1-x^k}}$. Find $k-m$ ?

Answer 1

Let $y={\cos }^{-1}(4x^3-3x)$
Put $x=\cos \theta$
$\Rightarrow y={\cos }^{-1}\cos 3\theta =3\theta =3{\cos }^{-1}x$
$\therefore \frac{dy}{dx}=-\frac{3}{\sqrt{1-x^2}}$