# How do I solve this equation?

## If $f \left(x\right) = {\left(x - 1\right)}^{2} \sin x$, then f'(0)=?

Apr 15, 2018

$f ' \left(0\right) = 1$

#### Explanation:

we have to use the product rule

$f \left(x\right) = \textcolor{red}{u} v \implies f ' \left(x\right) = \textcolor{red}{u '} v + \textcolor{red}{u} v '$

$f \left(x\right) = {\left(x - 1\right)}^{2} \sin x$

$\textcolor{red}{u = {\left(x - 1\right)}^{2} \implies u ' = 2 \left(x - 1\right)}$

$v = \sin x \implies v ' = \cos x$

$\therefore f ' \left(x\right) = \textcolor{red}{2 \left(x - 1\right)} \sin x + \textcolor{red}{{\left(x - 1\right)}^{2}} \cos x$

$\therefore f ' \left(0\right) = \cancel{2 \left(0 - 1\right) \sin 0} + {\left(0 - 1\right)}^{2} \cos 0$

$f ' \left(0\right) = 1 \times 1 = 1$