How do you differentiate #g(x) = e^(x-x^2)ln(3-x)# using the product rule?

1 Answer
Feb 10, 2016

Answer:

#g'(x)=(1-2x)e^(x-x^2)ln(3-x)+e^(x-x^2)/(x-3)#

Explanation:

The product rule shows us that

#g'(x)=ln(3-x)d/dx(e^(x-x^2))+e^(x-x^2)d/dx(ln(3-x))#

Both of these derivatives require use of the chain rule. We will make use of the following chain rule "identities":

#d/dx(e^x)=e^x" "=>" "d/dx(e^(f(x)))=e^(f(x)) * f'(x)#

#d/dx(ln(x))=1/x" "=>" "d/dx(ln(f(x)))=1/(f(x))*f'(x)#

So, we see that

#d/dx(e^(x-x^2))=e^(x-x^2)d/dx(x-x^2)=(1-2x)e^(x-x^2)#

and

#d/dx(ln(3-x))=1/(3-x)d/dx(3-x)=1/(3-x)(-1)=1/(x-3)#

Plug these both back in to find the function's derivative:

#g'(x)=(1-2x)e^(x-x^2)ln(3-x)+e^(x-x^2)/(x-3)#