# Calculus

posted by on .

I finished this problem but I wanted some feedback because I don't know if it's entirely correct.

Suppose that functions f and g and their derivatives with respect to x have the following values at x=2.
f(x)=8
g(x)=2
f'(x)=1/3
g'(x)=-3

Find d/dx of f(g(x)) at x=2.
I know the chain rule is this:
f'(g(x)) * g'(x)
So you would write the derivative of the outer function, keeping the inner function (g(x)) as the 'x' value. But in this case, the derivative of the outer function is a number.... I wasn't sure how to approach this problem. How do I figure in the g(x)?

This is the answer (and work) that I got:
f'(g(x)) * g'(x)
1/3 * -3 = -1
Where 1/3 is f'(x) and -3 is g'(x). But I feel like this is incorrect, because I didn't input the value of g(x).

Does anyone know if what I did was right, or how to correct it if it was wrong? I wasn't given an equation for f(x) or g(x) so I assumed we were supposed to rely on the numbers given. Do I need to multiply f'(x) and g(x) to get 2/3?

Thank you

• Calculus - ,

f'(g(x))* g'(x)
g(2) = 2
f'(2) = 1/3
so
f'(g(2)) = 2/3
and
g'(2) = -3
so I get
(2/3)(-3) = -2

• Calculus - ,

Thank you! That makes more sense... because you would say f' of g of x.... so 1/3 of 2 = 2/3. That makes sense!