Question

What is the derivative of $\sin x\cos x$?

Answer 1

Find the derivative of $\sin x\cos x$.

Let, $y=\sin x\cos x$

Differentiate both sides w.r.t $x$, using product rule.

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\sin x\cos x \\ =\sin x\frac{d}{dx}\cos x+\cos x\frac{d}{dx}\sin x\left[\because \frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}\right] \\ =\sin x\times \left(-\sin x\right)+\cos x\times \left(\cos x\right)\left[\begin{array}{c}\because \frac{d}{dx}\cos x=\sin x\\ \frac{d}{dx}\sin x=\cos x\end{array},\right] \\ =-{\sin }^{2}x+{\cos }^{2}x \\ ={\cos }^{2}x-{\sin }^{2}x \\ =\cos 2x\left[\because {\cos }^{2}x-{\sin }^{2}x=\cos 2x\right] \end{gathered}$$

Therefore, the derivative of $\sin x\cos x$ is $\cos 2x$.