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Chapter 5: Problem 41

In Exercises \(33-54,\) find the derivative of the function. $$y=x^{2} e^{-x}$$

Short Answer

Expert verified
\(y'=-x^{2}e^{-x}+2xe^{-x}\)

Step by step solution

01

Identify the two functions

We have \(f(x)=x^{2}\) and \(g(x)=e^{-x}\). We are going to find the derivatives of each of these functions.
02

Derivative of \(f(x)\)

The derivative of \(f(x)\) is \(f'(x)=2x\). This is using the power rule for derivatives that states that the derivative of \(x^n\) is \(nx^{n-1}\).
03

Derivative of \(g(x)\)

The derivative of \(g(x)\) is \(g'(x)=-e^{-x}\). The exponential function \(e^{x}\) differentiates to itself. But, there is a negative sign in the exponential. So, we have to apply chain rule here which states that the derivative of a function of a function is the derivative of the outer function times the derivative of the inner function so we get \(g'(x)=-e^{-x}\).
04

Apply the product rule

The product rule of differentiation states that the derivative of the product of two functions is the first function times the derivative of the second, plus the second function times the derivative of the first. Using our functions and their derivatives, we get \(y'=f(x)g'(x)+g(x)f'(x)=x^{2}\times (-e^{-x})+e^{-x}\times 2x=-x^{2}e^{-x}+2xe^{-x}\).

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