# Derivative help

1. Homework Statement
This problem is not hard at all its just that this stupid online homework program is a problem. Anyways,

f(x)=1/(5x+7)

The quotient:
f(7+h)+f(7)\h

This can be simplified to:
1\(A+Bh)

What is A & B
What is f'(7)?

2. Homework Equations

$$lim_{h\rightarrow0}\frac{\f(7+h)-f(7)}{h}$$

3. The Attempt at a Solution

The derivative is easy to get.
The final form is $$\frac{-5}{210h+1764}$$
How do I get it to:

$$\frac{1}{A+Bh}[\tex] Last edited: ## Answers and Replies Related Calculus and Beyond Homework Help News on Phys.org cristo Staff Emeritus Science Advisor Why not multiply top and bottom by -1/5? but when I input B=210 and A=1764 it tells me I'm incorrect cristo Staff Emeritus Science Advisor Well, you haven't multiplied them by -1/5. In order to get [tex]\frac{-5}{210h+1764}$$ into the required form, you need to multiply top and bottom by -1/5 (to get unity in the numerator).

Ah..of coarse.
There is another problem that requires the form Ah^2 + B h + C.
f(x) = 2x^2 + 9 x + 4, find f'(2).

Now I got :
2h+17, A is 0, B is 2 but
when i input 17 for c it is incorrect. Why?

Last edited:
Gib Z
Homework Helper
Umm do you have to do it by first principles? It seems the quotient rule would work fine here.

For your 2nd problem, f'(2)=17, I dont see what the problem is.

Umm do you have to do it by first principles? It seems the quotient rule would work fine here.

For your 2nd problem, f'(2)=17, I dont see what the problem is.
Well the question wants the form Ah^2+Bh+C
I know that
A=0
B=2
Shouldnt C=17? When i input that it is wrong.

If not c=17, I don't understand what else it could be!