# How do you differentiate g(x) =e^(1-x)sinx using the product rule?

Mar 10, 2016

$g ' \left(x\right) = f g ' + g f ' = {e}^{1 - x} \cos x - {e}^{1 - x} \sin x$

#### Explanation:

$f = {e}^{1 - x} , g = \sin x$
$f ' = {e}^{1 - x} \cdot - 1 = - {e}^{1 - x} , g ' = \cos x$
$g ' \left(x\right) = f g ' + g f ' = {e}^{1 - x} \cos x - {e}^{1 - x} \sin x$

Mar 10, 2016

${e}^{1 - x} \left(\cos x - \sin x\right)$

#### Explanation:

differentiate using the$\textcolor{b l u e}{\text{ Product rule }}$

If g(x)=f(x).h(x) then g'(x) = f(x).h'(x) + h(x).f'(x)
$\text{-------------------------------------------------------------}$
f(x) = ${e}^{1 - x} \Rightarrow f ' \left(x\right) = e \left(1 - x\right) \frac{d}{\mathrm{dx}} \left(1 - x\right) = - {e}^{1 - x}$

and h(x)=sinx $\Rightarrow h ' \left(x\right) = \cos x$
$\text{----------------------------------------------------------------}$
substitute these results into g'(x)

$\Rightarrow g ' \left(x\right) = {e}^{1 - x} . \cos x + \sin x . \left(- {e}^{1 - x}\right)$

taking out a common factor

$\Rightarrow g ' \left(x\right) = {e}^{1 - x} \left(\cos x - \sin x\right)$