Question

find the derivative of the function.
y = 5e^{-x}+e^{4x}
\frac{dy}{dx}=\square

Explanation:

Step1: Differentiate \(5e^{-x}\)

Use the chain rule \((e^{u})^\prime = e^{u}\cdot u^\prime\). Let \(u = -x\), then \(u^\prime=-1\). So \((5e^{-x})^\prime=5e^{-x}\cdot(-1)=-5e^{-x}\)

Step2: Differentiate \(e^{4x}\)

Let \(u = 4x\), then \(u^\prime = 4\). By the chain rule \((e^{4x})^\prime=e^{4x}\cdot4 = 4e^{4x}\)

Step3: Sum the derivatives

\(\frac{dy}{dx}=(5e^{-x})^\prime+(e^{4x})^\prime\)

Answer:

\(-5e^{-x}+4e^{4x}\)