Find the value

Question:

Find $\frac{d y}{d x}$, When $y=\sin \sqrt{\sin x+\cos x}$

Solution:

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

Formula : $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$ and $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$

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

According to the chain rule of differentiation

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

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

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