Question

Find the derivative of the following function:
f(x) = $(x^2+1)cosx$

Answer 1

Here $f\left(x\right)=(x^2+1)cosx$

$\therefore f^{\prime }\left(x\right)=\frac{d}{dx}\left[\right(x^2+1\left)cosx\right]$

$=(x^2+1)\frac{d}{dx}\left(cosx\right)+cosx\frac{d}{dx}(x^2+1)$

$=(x^2+1)(-sinx)+cosx\left(2x\right)$

$=-x^{2}sinx-sinx+2xcosx$