Question

$f\left(x\right)=x\cos \left(x\right)$. Find $f\text{'}\left(\mathrm{\pi }\right)$

Answer 1

Step 1: Differentiate the given function :

Given, $f\left(x\right)=x\cos \left(x\right)$

We know that, $\frac{d}{dx}\left(U.V\right)=U\frac{dV}{dx}+V\frac{dU}{dx}$

$$\begin{gathered} \therefore \frac{d}{dx}\left(x\cos \left(x\right)\right)=x\frac{d}{dx}\left(\cos \left(x\right)\right)+\cos \left(x\right)\frac{d}{dx}\left(x\right) \\ =x\left(-\sin \left(x\right)\right)+\cos \left(x\right)\left[\because \frac{d}{dx}\cos x=-\sin x;\frac{d}{dx}\left(x\right)=1\right] \\ \therefore f\text{'}\left(x\right)=\cos \left(x\right)-x\sin \left(x\right) \end{gathered}$$

Step 2: Substitute $\mathrm{\pi }$ into the derivative obtained.

$$\begin{gathered} \Rightarrow f\text{'}\left(\mathrm{\pi }\right)=\cos \left(\mathrm{\pi }\right)-\mathrm{\pi sin}\left(\mathrm{\pi }\right) \\ =-1-\mathrm{\pi }\left(0\right)\left[\because \cos \left(\mathrm{\pi }\right)=-1;\sin \left(\mathrm{\pi }\right)=0\right] \\ \therefore \mathrm{f}\text{'}\left(\mathrm{\pi }\right)=-1 \end{gathered}$$

Hence, the value of $f\text{'}\left(\mathrm{\pi }\right)$ is $-1$