Derivatives and the chain rule

[AL107]

Homework Statement

x: -1 1 3
f(x): 6 3. 1
f’(x): 5. -3 -2
g(x): 3. -1. 2
g’(x): -2. 2. 3

The table above gives values of f, f', g, and g' at selected values of x. If h(x) = f(g(x)), then h'(1) =
(A) 5
(B) 6
(C) 9
(D) 10
(E) 12

Relevant Equations

h(x)=f(g(x))

I originally thought you’d have to use the chain rule to get h’, as in: f’(g(x))*g’(x). Plugging in 1 for x, I got an answer of 10. An online solution, however, said that you only had to get f(g(1)), which was f(-1), then look up f’(-1) in the table. Both approaches seem logical to me, but they yield different results. Can someone clarify? Thank you!

 

[AL107]

Oh, here is a better table:

 

[PeroK]

An online solution, however, said that you only had to get f(g(1)), which was f(-1), then look up f’(-1) in the table.

How can that possibly be right? The chain rule applies.

 

[AL107]

How can that possibly be right? The chain rule applies.

Thank you!

 

[mjc123]

In the second method, you are differentiating h by x and f by g, so you are not performing the same operation on both sides of the equation. Be careful when using the prime notation: h'(x) means dh/dx, but f'(g) means df/dg. It may be helpful to write out the derivatives explicitly:
h(x) = f(g)
dh/dx = df/dx = df/dg*dg/dx

 

[FactChecker]

Intuitively, you can not just look at f' because x has to go through g before f is applied. Consider the simple example, f(x)=x. Then h(x)=f(g(x)) = g(x) and clearly h'= g', so g' can not be ignored.