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Formula: $\cos (A + B) = \cos A\cos B - \sin A\sin B$

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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:

$f'(x) = \dfrac{{df(x)}}{{dx}} = {f_2}(x).\dfrac{{d{f_1}(x)}}{{dx}} + {f_1}(x).\dfrac{{d{f_2}(x)}}{{dx}}$.

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:

$f'(x) = \dfrac{{df(x)}}{{dx}} = {f_2}(x).\dfrac{{d{f_1}(x)}}{{dx}} + {f_1}(x).\dfrac{{d{f_2}(x)}}{{dx}}$.