Question

find the derivative of f(x)= (e^ax)*(cos(bx)) using chain rule

Answer 1

If

$f(x) = e^{ax}\cos(bx)$

then by the product rule,

$f'(x) = \left(e^{ax}\right)' \cos(bx) + e^{ax}\left(\cos(bx)\right)'$

and by the chain rule,

$f'(x) = e^{ax}(ax)'\cos(bx) - e^{ax}\sin(bx)(bx)'$

which leaves us with

$f'(x) = \boxed{ae^{ax}\cos(bx) - be^{ax}\sin(bx)}$

Alternatively, if you exclusively want to use the chain rule, you can carry out logarithmic differentiation:

$\ln(f(x)) = \ln(e^{ax}\cos(bx)} = \ln(e^{ax})+\ln(\cos(bx)) = ax + \ln(\cos(bx))$

By the chain rule, differentiating both sides with respect to x gives

$\dfrac{f'(x)}{f(x)} = a + \dfrac{(\cos(bx))'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a - \dfrac{\sin(bx)(bx)'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a-b\tan(bx)$

Solve for f'(x) yields

$f'(x) = e^{ax}\cos(bx) \left(a-b\tan(bx)\right) \\\\ f'(x) = e^{ax}\left(a\cos(bx)-b\sin(bx))$

just as before.