Chain Rule Help: Solving (g o f)'(4) with f'(8)=5, g'(8)=3, f(4)=8, and g(4)=10

[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.

 

[AKG]

What have you tried; where are you getting stuck?

 

[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)

 

[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).

 

[ussjt]

what about this type of problem:

For a given functionhttps://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image1.png consider[/URL] the composite function https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image2.png Suppose we know that https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image3.png

Calculate f ' (x)

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

 

[HallsofIvy]

what about this type of problem:

For a given functionhttps://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image1.png consider[/URL] the composite function https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image2.png Suppose we know that https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk5/696680/kopko.4-prob5image3.png

Calculate f ' (x)

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

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)?

 

[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)
~~~~~~~~~~

answer: (7x^3/6)*(y/2)^(1/3)

 

[VietDao29]

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

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?

 

[ussjt]

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?

where did the f'(y) come from?

 

[NateTG]

where did the f'(y) come from?

In VietDao's post y is a place holder for [itex]2x^3[/itex]

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

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

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