Question

find the derivative of the function.
f(t) = e^{mt} cos(nt)

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = e^{mt}\) and \(v=\cos(nt)\).
First, find \(u^\prime\):
Using the chain rule, if \(u = e^{mt}\), then \(u^\prime=\frac{d}{dt}(e^{mt})=me^{mt}\) (since \(\frac{d}{dx}(e^{ax}) = ae^{ax}\)).
Next, find \(v^\prime\):
Using the chain rule, if \(v=\cos(nt)\), then \(v^\prime=\frac{d}{dt}(\cos(nt))=-n\sin(nt)\) (since \(\frac{d}{dx}(\cos(ax))=-a\sin(ax)\)).

Step2: Substitute into the product rule

\(f^\prime(t)=u^\prime v+uv^\prime\)
\(f^\prime(t)=me^{mt}\cos(nt)+e^{mt}(-n\sin(nt))\)
Factor out \(e^{mt}\):
\(f^\prime(t)=e^{mt}(m\cos(nt)-n\sin(nt))\)

Answer:

\(e^{mt}(m\cos(nt)-n\sin(nt))\)