Question

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

Answer 1

`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)`