Question

If f(x) = x.cos(x) then find the value of $f'(\pi )$.
Formula: $\cos (A + B) = \cos A\cos B - \sin A\sin B$

Answer 1

Hint- We will be using the differentiation of $f(x) = x\cos x$ and here we are differentiating two functions of $x$ with respect to $x$. Then replace $x$ by $\pi $ to obtain $f'(\pi )$ as shown below. Use $\cos \pi = - 1$ and $\sin \pi = 0$.

Complete step-by-step answer:
Given that, $f(x) = x\cos x$
Now differentiating with respect to $x$, we get
$ \Rightarrow f'(x) = \dfrac{{df(x)}}{{dx}} = \cos x\dfrac{{d(x)}}{{dx}} + x\dfrac{{d(\cos x)}}{{dx}}$
$ \Rightarrow f'(x) = \cos x - x\sin x$
$ \Rightarrow f'(\pi ) = \cos \pi - \pi \sin \pi \Rightarrow - 1 - \pi .0$
$ \Rightarrow f'(\pi ) = - 1$

Note- Here, the $f(x) = x\cos x$ is a product of two functions of $x$. We have considered $f(x) = {f_1}(x).{f_2}(x)$ where ${f_1}(x) = x$ and ${f_2}(x) = \cos x$. We have used differentiation formula for product of two functions as shown below:
$f'(x) = \dfrac{{df(x)}}{{dx}} = {f_2}(x).\dfrac{{d{f_1}(x)}}{{dx}} + {f_1}(x).\dfrac{{d{f_2}(x)}}{{dx}}$.