Question

If f(x) = x.cos(x) then find the value of $f^{\prime }(\pi )$.
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 $f^{\prime }(\pi )$ 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
$\Rightarrow f^{\prime }(x)={\displaystyle \frac{df(x)}{dx}}=\cos x{\displaystyle \frac{d(x)}{dx}}+x{\displaystyle \frac{d(\cos x)}{dx}}$
$\Rightarrow f^{\prime }(x)=\cos x-x\sin x$
$\Rightarrow f^{\prime }(\pi )=\cos \pi -\pi \sin \pi \Rightarrow -1-\pi .0$
$\Rightarrow f^{\prime }(\pi )=-1$

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:
$f^{\prime }(x)={\displaystyle \frac{df(x)}{dx}}=f_2(x).{\displaystyle \frac{d{f}_1(x)}{dx}}+f_1(x).{\displaystyle \frac{d{f}_2(x)}{dx}}$.