Question

If $y=\sin \left[\sqrt{\sin \sqrt{x}}\right]$, then find $\frac{dy}{dx}$.

Answer 1

$y=\sin \left(\sqrt{\sin \sqrt{x}}\right)$

From chain rule -

$\Rightarrow \frac{dy}{dx}=\frac{dy}{d\sqrt{\sin \sqrt{x}}}\times \frac{d\sqrt{\sin \sqrt{x}}}{d\sqrt{x}}\times \frac{d\sin \sqrt{x}}{d\sqrt{x}}\times \frac{d\sqrt{x}}{dx}$

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

$=\frac{1}{4}\cos \sqrt{\sin \sqrt{x}}{\left(x\sin \sqrt{x}\right)}^{-1/2}\cos \sqrt{x}.$

Hence, solved.