AP Calculus BC - Chain Rule

Question

If f(3) = 4, g(3) = 2, f'(3) = -5, g'(3) = 6, find the following:

a. (f + g)'(3)

b. (f∘g²)'(3) --> (function composition)

Answer 1

Hi Washington,

For part (a), (f+g)'(x)= f'(x)+g'(x). Let x=3:

f'(3)+g'(3)= -5+6=1

For part (b), (f_°g²)'(x)= (d/dx)(f(g²(x)))

*NOTICE AN OPEN CIRCLE INDICATES IT'S A COMPOSITION OF FUNCTIONS, NOT MULTIPLICATION WHICH IS A CLOSED CIRCLE! (As I'm sure you know)

We have a function in a function in a function. So it's a double chain rule. Differentiate from the "outside in":

d/dx= f'(g²(x))*2(g(x))*g'(x)

Plug 3 in and we have: f'(4)*2(2)*6= 24(f'(4)).

f'(4) is not given.

I hope this helps!