Question

If $y=\sin \left(\sqrt{\sin x+\cos x}\right)$, then $\frac{dy}{dx}=$

• A. $\frac{\cos \left(\sqrt{\sin x+\cos x}\right)\left(\cos x-\sin x\right)}{2\sqrt{\sin x+\cos x}}$
• B. $\frac{\cos \left(\sqrt{\sin x+\cos x}\right)}{2\sqrt{\sin x+\cos x}}$
• C. $\frac{\cos \left(x\sqrt{\sin x+\cos x}\right)\left(\cos x-\sin x\right)}{2\sqrt{\sin x-\cos x}}$
• D. none of these.

Answer 1

The correct option is A

$\frac{\cos \left(\sqrt{\sin x+\cos x}\right)\left(\cos x-\sin x\right)}{2\sqrt{\sin x+\cos x}}$

The explanation for the correct option:

The given equation is $y=\sin \left(\sqrt{\sin x+\cos x}\right)$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left[\sin \left(\sqrt{\sin x+\cos x}\right)\right] \\ \Rightarrow \frac{dy}{dx}=\cos \left(\sqrt{\sin x+\cos x}\right)·\frac{d}{dx}\left(\sqrt{\sin x+\cos x}\right) \\ \Rightarrow \frac{dy}{dx}=\cos \left(\sqrt{\sin x+\cos x}\right)·\left(\frac{1}{2}\right)·{\left(\sin x+\cos x\right)}^{\frac{1}{2}-1}·\frac{d}{dx}\left(\sin x+\cos x\right) \\ \Rightarrow \frac{dy}{dx}=\cos \left(\sqrt{\sin x+\cos x}\right)·\left(\frac{1}{2}\right)·{\left(\sin x+\cos x\right)}^{-\frac{1}{2}}·\left(\cos x-\sin x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{\cos \left(\sqrt{\sin x+\cos x}\right)\left(\cos x-\sin x\right)}{2\sqrt{\sin x+\cos x}} \end{gathered}$$

Therefore, $\frac{dy}{dx}=\frac{\cos \left(\sqrt{\sin x+\cos x}\right)\left(\cos x-\sin x\right)}{2\sqrt{\sin x+\cos x}}$.

Hence, (A) is the correct option.