# Chain rule help

1. Feb 12, 2006

### ussjt

f '(8)=5 g '(8)=3 f(4)=8 g(4)=10 g(4)=10 g(8)=2 f(8)=5

find (g o f)'(4)

how do I go about setting up these types of problem.

2. Feb 12, 2006

### AKG

What have you tried; where are you getting stuck?

3. Feb 12, 2006

### ussjt

have not tried because I don't know how to set up the problem...I don't really care about the answer I just want to know how you go about setting up these kinds of problems because I have a quiz tomorrow.

I figure the first step is g(f(x))

so g'(f(x))*f'(x)
g'(8)*f'(x)
3*f'(x)

Last edited: Feb 12, 2006
4. Feb 12, 2006

### AKG

(g o f)'(4)
= g'(f(4))*f'(4) by chain rule
= g'(8)*f'(4) since f(4) = 8
= 3*f'(4) since g'(8) = 3

And that's all you can do, since they don't tell you what f'(4) is. I suspect they do, and you just copied out the question wrong. Also, why have you given "g(4) = 10" twice?

Anyways, the way to setting up the problem is this:

Given a problem, "find X", write:

X
= A (by some theorem, or given fact, or logical inference)
= B (again, give justification)
= C (justification)
= D (justification)

until you get some answer D that you think the teacher will like, like an actual numeral. In this case, your X is (f o g)'(4), and your C is something like 3f'(4). You want a numeral for your D, but you can't get it yet from C because they haven't given you enough information (or you copied the question wrong).

5. Feb 12, 2006

### ussjt

For a given functionconsider[/PLAIN] [Broken] the composite function [Broken] Suppose we know that [Broken]

Calculate f ' (x)

How do I go about setting up this type of problem?

Last edited by a moderator: Apr 22, 2017 at 8:10 AM
6. Feb 13, 2006

### HallsofIvy

Staff Emeritus
You titled this thread "chain rule"! It ought to occur to you to use the chain rule!
If h(x)= f(2x3) then h'(x)= f '(2x3)(6x2).

You are given that h'(x)= 7x5.

You can easily solve f '(2x3)(6x2)= 7x5 for f '(2x3).

Now let y= 2x3. What is f(y)?

Last edited by a moderator: Apr 22, 2017 at 8:11 AM
7. Feb 14, 2006

### ussjt

the way our TA showed up, the answer ought to be:

(7x^3/6)*(y/2)^(1/3)...but my answer must be in terms of x...so could someone please tell me if I went wrong somewhere or how to make it all in terms of x (by x I mean I can't have that "y"). Here are my steps:

f '(2x^3)(6x^2)= 7x^5

f '(2x^3)= (7x^5)/(6x^2)

f '(2x^3)= (7x^3)/6
~~~~~~~~~~~~
2x^3=y

x^3= y/2

x= (y/2)^(1/3)
~~~~~~~~~~

8. Feb 15, 2006

### VietDao29

It's fine up to here.
Now sub what you get in the expression:
f '(2x3)= (7x3)/6, we have:
f'(y) = 7y / 12.
So what's f'(x)?
Can you go from here?

9. Feb 15, 2006

### ussjt

where did the f'(y) come from?

10. Feb 15, 2006

### NateTG

In VietDao's post y is a place holder for $2x^3$

I don't know if this will make things any clearer for you:

The problem gives you
$$h(x)=f(z(x))$$
So, let's say we have some $a$ so that $x=z^{-1}(a)$ (provided that $z^{-1}$ actually exists). Then we can substitute that in
$$h(z^{-1}(a))=f(z(z^{-1}(a))$$
then simplify
$$h(z^{-1}(a))=f(a)$$
Now, we can take the derivative of both sides w.r.t. a
$$h'(z^{-1}(a)) \times \left(z^{-1}\right)' (a) = f'(a)$$

Now, since $h'$ and $z$ are both known, you should be able to work out what the left hand side of the equation is equal to.