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

Question

Emma Aldridge · Accepted Answer

#f'(x) = e^(-x) (cos x - sin x)#

The following is the product rule: if your function is the result of two functions being combined,

#f(x) = g(x) * h(x)#

subsequently, the derivative is

#f'(x) = g'(x) * h(x) + g(x) * h'(x)#

In your case, #g(x) = e^(-x)# and #h(x) = sin x#.

Let's find the derivatives #g'(x)# and #h'(x)#:

#g(x) = e^(-x) color(white)(xxx) => color(white)(x) g'(x) = -e^(-x)#

#h(x) = sin x color(white)(xxi) => color(white)(x) h'(x) = cos x#

Thus, your derivative as a whole is:

#f'(x) = g'(x) * h(x) + g(x) * h'(x)#

#color(white)(xxxx) = -e^(-x) sin x + e^(-x) cos x#

#color(white)(xxxx) = color(white)(x) e^(-x) (cos x - sin x)#

Axel Alvarez · Answer

undefined To differentiate $ f(x) = e^{-x} \sin(x) $ using the product rule, you apply the formula:

$$ (uv)' = u'v + uv' $$

where $ u = e^{-x} $ and $ v = \sin(x) $.

Then, take the derivatives of $ u $ and $ v $:

$$ u' = -e^{-x} $$
$$ v' = \cos(x) $$

Now, apply the product rule:

$$ f'(x) = (e^{-x})(\sin(x))' + (e^{-x})'(\sin(x)) $$

$$ = (-e^{-x} \sin(x)) + (e^{-x} \cos(x)) $$

$$ = e^{-x} (\cos(x) - \sin(x)) $$