Question

How do you find the 1st and 2nd derivative of `y=x^2e^x-4xe^x+2e^x`?

Answer 1

`dy/dx= e^x(x^2-2x-2)` or `x^2e^x-2xe^x-2e^x`
`:. (d^2y)/dx^2= e^x(x^2-4)` or `x^2e^x-4e^x`

Explanation:

Use the product rule: `d/dxuv= u(dv)/dx+v(du)/dx`

So `y = x^2e^x-4xe^x+2e^x`
`:. y = (x^2-4x+2)e^x`

The product rule gives:

`dy/dx= (x^2-4x+2)d/dxe^x+e^xd/dx(x^2-4x+2)`
`:. dy/dx= (x^2-4x+2)e^x+e^x(2x-4)`
`:. dy/dx= e^x(x^2-4x+2+2x-4)`
`:. dy/dx= e^x(x^2-2x-2)`

Applying the product rule again:
`(d^2y)/dx^2 = (x^2-2x-2)d/dxe^x+e^xd/dx(x^2-2x-2)`
`:. (d^2y)/dx^2= (x^2-2x-2)e^x+e^x(2x-2)`
`:. (d^2y)/dx^2= e^x(x^2-2x-2+2x-2)`
`:. (d^2y)/dx^2= e^x(x^2-4)`