# Finding derivative

1. Oct 12, 2009

### stanton

1. The problem statement, all variables and given/known data

(i)f(x)/3=g(f)
(ii)The tangent to f(x) at x=2 (point P(2,3)) has equation y=2.1x-1
Find the value of f '(2)
Find the valus of g(2)
Find the value of g '(2)

2. Relevant equations

Chain rule, I guess?
d/dx x [f(g(x))]=f'(g(x))g'(x) I think this is it?
dy/dx=dy/du x du/dx

3. The attempt at a solution

What I want to know is that what does (ii)mean? does it mean line tangent to f(x) at x=2 is [ y=2.1x-1]? (in other words, the equation for slope of f(x) is y=2.1x-1 at x=2)
Or does it mean f(x) has an equation y=2.1x-1 at x=2?
So I had two solution to this prob.
1. regarding that f '(x) is y=2.1x-1, plug the value 2 in the equation, so the answer is 3.2
2. regarding that f(x) is y=2.1x-1 at x=2, f '(x) is 2.1. so the answer is 2.1
Help me. How should I interpret the question?

As I did not understan whether f(x) is y=2.1x-1 or equation of slope is y=2.1x-1, I am stuck now. But if f(x) is y=2.1x-1, I can plug 2 in g(x) like this to solve problem a:
2.1x-1/3=2.1(2)/3=1.4

Last edited: Oct 12, 2009
2. Oct 12, 2009

### darkchild

(ii) means that the line that is tangent to the graph of f(x) at the point x=2 has the given equation. To solve, you need to understand the connection between tangent lines to graphs and the derivative.

3. Oct 12, 2009

### Staff: Mentor

Shouldn't this be f(x)/3 = g(f(x))? Also, what are you asked to find?
It means the first possibility. The tangent line to the graph of y = f(x) has this equation: y = 2.1x -1.
Yes, but what you want to say is that f(2) = 3.2.
Since f(x)/3 = (1/3)f(x) = g(f(x)), can you describe what function g does to an input value? Knowing that will enable you to write a formula for g(x), and will enable you to do the 2nd part.
Yes, this is what you want, but it's not "d/dx" times [f(g(x))]; it's d/dx of [f(g(x))]. You're taking the derivative, with respect to x, of [f(g(x))]. d/dx is not a number, so it makes no sense to talk about multiplying something by it.

Your chain rule formula should look like this: d/dx[f(g(x))]=f'(g(x))g'(x)

4. Oct 12, 2009

### stanton

Thank you so much. I was very confused. So f'(2) is 3.2, right? For f'(x) at x=2 is 2.1x-1
Also , I figured out the f(x). I fogot to enter coordinates in my post when x=2 :(2,3)
I pluged this in equation y=ax+b. So 3=2a+b
And the slope is given as 3.2
f(x) has this equation at x=2: y=3.2x-3.4 So g(x) is y=3.2x-3.4/3
Therefore, g(x)=1.06x-1.1
I pugged the value 2 in g(x) and got 2.2 And I got g' (x)=-1.1
And g(2)=1.02
Am I doing right so far?
And Thank you for your all the advices and help. I really appreciate it.

Last edited: Oct 12, 2009
5. Oct 12, 2009

### Staff: Mentor

Right. I inadvertently omitted the '. I meant to say that f'(2) = 3.2.
No, f'(x) at x = 2 (i.e., f'(2)) is 2.1*2 - 1 = 3.2, not 2.1x - 1.
You're way off here. I assume you're talking about the problem where you need to find g(2) and g'(2), right?

You need to find the formula for g(x) from f(x)/3 = g(f(x)). My question again is what does g do to an input value? You need to answer this before attempting to find g(2) and g'(2).

6. Oct 12, 2009

### stanton

I think I went in the wrong way with solving g(2) and g'(2). I reinterpreted the question. Could you check this for me?
y=2.1x-1
f' (x)=2.1
If we integrate f' (x) with respect to x,
f(x)=2.1x+C
Plug coordinate(2,3)--> f(x)=2.1x-1.2 g(x)=0.7x-0.4
So g(2) is 1. g' (x)=0.7
And sorry, to bother you, I can't understand your question. I don't see any role of g. :(

7. Oct 12, 2009

### Staff: Mentor

This is what I was working with:
It complicates things when you post more than one question in a post. Also, when the problem is spread out all over the post. Use the first section for "the problem statement, all variables and given/known data." No part of the problem statement should be in the attempt at a solution section.

So what exactly is the problem you're trying to solve?

8. Oct 12, 2009

### stanton

I fixed my post :) In first section(problem statement) I put my problems that I am trying to solve.

9. Oct 12, 2009