Differentiate the following function with respect to x.
$\left( 1+{{x}^{2}} \right)\cos x.$

Hint: when we have to differentiate product of two continuous functions we use$\dfrac{d}{dx}\left[ f(x)g(x) \right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}g(x)$

Complete step-by-step answer:
Now let us assume here $f(x)=1+{{x}^{2}}$ and $g(x)=\cos x$
So we can write$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left[ (1+{{x}^{2}})\cos x \right]=(1+{{x}^{2}})\dfrac{d}{dx}(\cos x)+\cos x\dfrac{d}{dx}(1+{{x}^{2}}) \\
 & \Rightarrow \dfrac{d}{dx}\left[ (1+{{x}^{2}})\cos x \right]=(1+{{x}^{2}})(-\sin x)+\cos x(0+2x) \\
\end{align}$
As we know that
$\begin{align}
  & \dfrac{d}{dx}\cos x=-\sin x \\
 & \dfrac{d(cons\tan t)}{dx}=0 \\
 & \dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}} \\
 & \dfrac{d}{dx}\left[ f(x)+g(x) \right]=\dfrac{d}{dx}\left[ f(x) \right]+\dfrac{d}{dx}\left[ g(x) \right] \\
\end{align}$
So we can write
$\dfrac{d}{dx}\left[ (1+{{x}^{2}})\cos x \right]=-\sin x-{{x}^{2}}\sin x+2x\cos x$

Note: Here we can take $g(x)=(1+{{x}^{2}})$ and$f(x)=\cos x,$ the result will be the same. Order is not important when we differentiate the product of two continuous functions.