If #f(x)= 2 x^2 - 3 x # and #g(x) = 2e^x + 1 #, how do you differentiate #f(g(x)) # using the chain rule?

2 Answers

Answer:

Hi there!

It looks like you're being asked to differentiate a composite function! You'll want to start off by subbing g(x) into f, then you can differentiate.

Explanation:

Starting off with composing the function, or subbing g(x) into f. Take g(x) and sub it into wherever there is an x:

#f(g(x)) = 2(2e^x+1)^2-3(2e^x+1) #

Now you can differentiate using the chain rule. Let's differentiate:

Remember that when differentiating something of the form:

# f(x) = a(g(x))^n#, by the chain rule you get:

#f'(x) = an(g(x))^(n-1) * g'(x) #

Now, using the same principle you can differentiate:

#f(g(x))' = 4(2e^x+1)^1*2e^x -3(2e^x+1)^0*2e^x #

Simplifying we get:

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

And that's pretty much it! Most teachers/professors are okay with leaving it unsimplified, as cleaning it up doesn't change the answer.

Hopefully you've understood how the chain rule was applied! If you have any questions, let me know!

Mar 29, 2016

Answer:

Using a less formal notation on part only as a demonstration

Explanation:

Consider : # 2(2e^x+1)^2#

Let #u=2e^x+1 # then #(du)/(dx) = 2e^x#

Let # y=2u^2 # then #(dy)/(du) = 4u#

But #(dy)/(dx) = (dy)/(du)xx(du)/(dx)#

so #(dy)/(dx) = 4uxx2e^x" "=" "4(2e^x+1)xx2e^x#

so #(dy)/(dx) = 8e^x(2e^x+1)#

Which can be worked on further

This is only a demonstration of method