Question

Find the derivative of ${\sin }^{2}x$ with respect to $x$using product rule.

Answer 1

Find the derivative using the product rule:

Given function : ${\sin }^{2}x$

$\Rightarrow \sin x\times \sin x$

Let ,$u=\sin x$ and $v=\sin x$

We know that the derivative of $u\times v$ as per the product rule of differentiation is $\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}$ ,where ${\displaystyle \frac{d}{dx}}\left(uv\right)$ is the derivative of $uv$ with respect to$x$ ,${\displaystyle \frac{dv}{dx}}$ is the derivative of $v$ with respect to$x$ and ${\displaystyle \frac{du}{dx}}$ is the derivative of $u$ with respect to$x$.

By substituting the value in above formula we get,

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left({\sin }^{2}x\right)=\sin x\frac{d}{dx}\left(\sin x\right)+\sin x\frac{d}{dx}\left(\sin x\right) \\ \Rightarrow \frac{d}{dx}\left({\sin }^{2}x\right)=\sin x\cos x+\sin x\cos x \\ \Rightarrow \frac{d}{dx}\left({\sin }^{2}x\right)=2\sin x\cos x \\ \Rightarrow \frac{d}{dx}\left({\sin }^{2}x\right)=\sin 2x\left[\because 2\sin x\cos x=\sin 2x\right] \end{gathered}$$

Hence, the derivative of ${\sin }^{2}x$ with respect to $x$is $\sin 2x$.