How do you find the 1st and 2nd derivative of y=x^2e^(x^2)?

1 Answer
mason m
Nov 5, 2016

dy/dx=2xe^(x^2)(1+x^2) and (d^2y)/dx^2=2e^(x^2)(2x^4+5x^2+1)

Explanation:

We start by finding the first derivative through the product rule, which says that d/dx(uv)=(du)/dx*v+u*(dv)/dx. Thus:

dy/dx=(d/dxx^2)e^(x^2)+x^2(d/dxe^(x^2))

We see that:

  • d/dxx^2=2x
  • d/dxe^(x^2)=e^(x^2)(d/dxx^2)=e^(x^2)(2x)

Note that you need to use the chain rule for the derivative of e^(x^2). We need to remember that d/dxe^x=e^x, so d/dxe^u=e^u*(du)/dx.

Returning to the derivative:

dy/dx=(2x)e^(x^2)+x^2(e^(x^2))(2x)

dy/dx=2xe^(x^2)+2x^3e^(x^2)

dy/dx=2e^(x^2)(x+x^3)

Note that we can also write that dy/dx=2xe^(x^2)(1+x^2) but this will make differentiating more difficult.

Now to find the second derivative, use the product rule again!

(d^2y)/dx^2=2(d/dxe^(x^2))(x+x^3)+2e^(x^2)(d/dx(x+x^3))

We already know that d/dxe^(x^2)=e^(x^2)(2x). Through the product rule, d/dx(x+x^3)=1+3x^2. So:

(d^2y)/dx^2=2(e^(x^2))(2x)(x+x^3)+2e^(x^2)(1+3x^2)

(d^2y)/dx^2=e^(x^2)(4x^2+4x^4)+e^(x^2)(2+6x^2)

(d^2y)/dx^2=e^(x^2)(4x^4+10x^2+2)

(d^2y)/dx^2=2e^(x^2)(2x^4+5x^2+1)

Related questions
See all questions in Differentiating Exponential Functions with Base e
Impact of this question
6315 views around the world
You can reuse this answer
Creative Commons License