# How do you differentiate f(x)=(sinx+1)(x^2-3e^x) using the product rule?

May 23, 2016

$\setminus \cos \left(x\right) \left({x}^{2} - 3 {e}^{x}\right) + \left(2 x - 3 {e}^{x}\right) \left(\setminus \sin \left(x\right) + 1\right)$

#### Explanation:

$\setminus \frac{d}{\mathrm{dx}} \left(\left(\setminus \sin \left(x\right) + 1\right) \left({x}^{2} - 3 {e}^{x}\right)\right)$

Applying product rule,
${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$

$f = \sin x + 1 , g = {x}^{2} - 3 {e}^{x}$
$= \frac{d}{\mathrm{dx}} \left(\sin \left(x\right) + 1\right) \left({x}^{2} - 3 {e}^{x}\right) + \frac{d}{\mathrm{dx}} \left({x}^{2} - 3 {e}^{x}\right) \left(\sin \left(x\right) + 1\right)$

we know,
$\setminus \frac{d}{\mathrm{dx}} \left(\sin \left(x\right) + 1\right) = \cos \left(x\right)$;
$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \sin \left(x\right)\right) = \setminus \cos \left(x\right)$;
$\setminus \frac{d}{\mathrm{dx}} \left(1\right) = 0$;
$\setminus \frac{d}{\mathrm{dx}} \left({x}^{2} - 3 {e}^{x}\right) = 2 x - 3 {e}^{x}$

Finally,
$= \setminus \cos \left(x\right) \left({x}^{2} - 3 {e}^{x}\right) + \left(2 x - 3 {e}^{x}\right) \left(\setminus \sin \left(x\right) + 1\right)$