Question

How do you differentiate `f(x)=(-e^x+x)(3x^3-x)` using the product rule?

Answer 1

See below.

Explanation:

By the product rule `d/dxu(x)v(x)=u'(x)v(x)+v'(x)u(x)`
Therefore, if we let `u(x)=-e^x+x` and `v(x)=3x^3-x`
Then `d/dx(-e^x+x)(3x^3-x)=(-e^x+x)'(3x^3-x)+(3x^3-x)'(-e^x+x)`
`(-e^x+x)'=-e^x+1`
`(3x^3-x)'=9x^2-1`
Therefore, the derivative `=(-e^x+1)(3x^3-x)+(9x^2-1)(-e^x+x)`
You could then expand and simply to finish off.