Question

find the derivative of the function.
y = xe^{-6x}-3e^{x^{3}}
\frac{dy}{dx}=\square
(type an exact answer in terms of e.)

Explanation:

Step1: Differentiate \(xe^{-6x}\) using product rule

Product rule: \((uv)^\prime = u^\prime v+uv^\prime\). Let \(u = x\), \(u^\prime=1\); \(v = e^{-6x}\), \(v^\prime=-6e^{-6x}\). Then \((xe^{-6x})^\prime=e^{-6x}+x(-6e^{-6x})=(1 - 6x)e^{-6x}\)

Step2: Differentiate \(-3e^{x^{3}}\) using chain rule

Chain rule: \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(f(u)=-3e^{u}\), \(f^\prime(u)=-3e^{u}\); \(u = x^{3}\), \(u^\prime = 3x^{2}\). Then \((-3e^{x^{3}})^\prime=-3e^{x^{3}}\cdot3x^{2}=-9x^{2}e^{x^{3}}\)

Step3: Combine the derivatives

\(\frac{dy}{dx}=(1 - 6x)e^{-6x}-9x^{2}e^{x^{3}}\)

Answer:

\((1 - 6x)e^{-6x}-9x^{2}e^{x^{3}}\)