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Lynna [10]
3 years ago
11

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

Mathematics
1 answer:
Vikentia [17]3 years ago
4 0

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 <em>x</em> 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 <em>f'(x)</em> 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.

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