Differentiate w.r.t x:

Let $y=\cos (\sin \sqrt{a x+b}), z=\sin \sqrt{a x+b}$ and $w=\sqrt{a x+b}$

Formula :

$\frac{d(\cos x)}{d x}=-\sin x$ and $\frac{d(\sin x)}{d x}=\cos x$

$\frac{\mathrm{d}(\sqrt{\mathrm{ax}+\mathrm{b}})}{\mathrm{dx}}=\frac{1}{2} \times(\mathrm{ax}+\mathrm{b})^{\frac{1}{2}-1} \times \mathrm{a}$

According to the chain rule of differentiation

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{dz}} \times \frac{\mathrm{dz}}{\mathrm{dw}} \times \frac{\mathrm{dw}}{\mathrm{dx}}$

$=-\sin (\sin \sqrt{a x+b}) \times \cos \sqrt{a x+b} \times \frac{1}{2} \times(a x+b)^{-\frac{1}{2}} \times a$

$=-\frac{a}{2} \sin (\sin \sqrt{a x+b}) \times \cos \sqrt{a x+b} \times(a x+b)^{-\frac{1}{2}}$