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

1 Answer
Feb 7, 2017

Answer:

#g'(x) = (1-4x)e^(3-x) - (x-2x^2)e^(3-x)#

Explanation:

Assuming the function is actually #g(x)# and not #g(y)#, the following is the explanation. Otherwise, differentiating with respect to #y# means treating #x# as a constant, so #g'(y) = 0#.

The product rule states that for #a,b# functions of #x#,

#(ab)' = a'b + ab'#

If we let #a(x) = (x-2x^2)# and #b(x) = e^(3-x)#, then:

We know that #a'(x) = 1-4x# from the power rule.

For #b'(x)#, we also need to use the chain rule, which states that if #c,d# functions of #x#, then

#(c(d(x)))' = c'(d(x)) * d'(x)#

If we let #c(x) = e^x# and #d(x) = 3-x# then the composition of #c# with #d# will be #e^(3-x)#.

The derivative of #c# with respect to #d(x)# is #e^3-x# and the derivative of #d# with respect to #x# is #color(red)(-1)#.

We have to multiply with that #-1#, which is why we see the minus sign in between the terms in the answer.

So, #g'(x) = (1-4x)e^(3-x) - (x-2x^2)e^(3-x)#