Question

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

Answer 1

Let $x=\sin \theta$

then $1-x^2=1-{\sin }^2\theta ={\cos }^2\theta$

$y={\cos }^{-1}\{\frac{3}{5}\sin \theta +\frac{4}{5}\cos \theta \}$

Let $\cos A=\frac{4}{5},\sin A=\frac{3}{5}$

$\therefore y={\cos }^{-1}\{\cos A\cos \theta +\sin A\sin \theta \}={\cos }^{-1}\left\{\cos (\theta -A)\right\}=\theta -A$

Now, $\frac{dy}{d\theta }=1-\left(1\right)$

Also we have $x=\sin \theta$

$\Rightarrow \theta ={\sin }^{-1}x\Rightarrow d\theta =\frac{1}{\sqrt{1-x^2}}dx-\left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$ we get

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