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

1 Answer
Mar 6, 2016

Answer:

#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#