Question

differentiate. y = xe^(-4x)+5e^(-x)+x^8 choose the correct answer below. a. dy/dx = - 4xe^(-4x)+e^(-4x)-5e^(-x)+8x^7 b. dy/dx = - 4xe^(-4x)-e^(-4x)-5e^(-x)+8x^7 c. dy/dx = - 4xe^(-4x)+e^(-4x)+5e^(-x)+8x^7 d. dy/dx = - 4xe^(-4x)+e^(-4x)-5e^(-x)-8x^7

Explanation:

Step1: Apply product - rule on $xe^{-4x}$

The product - rule states that if $u = x$ and $v=e^{-4x}$, then $(uv)'=u'v + uv'$. Here, $u' = 1$ and $v'=-4e^{-4x}$. So, $(xe^{-4x})'=e^{-4x}-4xe^{-4x}$.

Step2: Differentiate $5e^{-x}$

Using the rule that if $y = ae^{bx}$, then $y'=abe^{bx}$, for $a = 5$ and $b=-1$, we have $(5e^{-x})'=-5e^{-x}$.

Step3: Differentiate $x^{8}$

Using the power - rule $(x^{n})'=nx^{n - 1}$, for $n = 8$, we get $(x^{8})'=8x^{7}$.

Step4: Combine the derivatives

$y'=(xe^{-4x})'+(5e^{-x})'+(x^{8})'=e^{-4x}-4xe^{-4x}-5e^{-x}+8x^{7}$.

Answer:

A. $\frac{dy}{dx}=-4xe^{-4x}+e^{-4x}-5e^{-x}+8x^{7}$