Question

How do you differentiate `f(x)=x-x^2e^x-xlnx` using the product rule?

Answer 1

`f'(x)=-2xe^x-x^2e^x-lnx`

Explanation:

The product rule states that

`d/dx[g(x)h(x)]=g'(x)h(x)+g(x)h'(x)`

When finding `f'(x)`, we will have to apply the product rule to both `-x^2e^x` and `-xlnx`.

We can find the derivative of each term individually:

`d/dx(x)=1`

`d/dx(-x^2e^x)=e^xd/dx(-x^2)+(-x^2)d/dx(e^x)`

`color(white)l=-2xe^x-x^2e^x`

`d/dx(-xlnx)=lnxd/dx(-x)+(-x)d/dx(lnx)`

`color(white)(ll)=lnx(-1)+(-x)1/x`

`color(white)(ll)=-lnx-1`

Combining all these terms, we see that the function's entire derivative is

`f'(x)=1-2xe^x-x^2e^x-lnx-1`

`f'(x)=-2xe^x-x^2e^x-lnx`