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

1 Answer
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)#