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Derivative help

  • Thread starter Weave
  • Start date
143
0
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

[tex]lim_{h\rightarrow0}\frac{\f(7+h)-f(7)}{h}[/tex]


3. The Attempt at a Solution

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

[tex]\frac{1}{A+Bh}[\tex]
 
Last edited:

Answers and Replies

cristo
Staff Emeritus
Science Advisor
8,056
72
Why not multiply top and bottom by -1/5?
 
143
0
but when I input B=210 and A=1764 it tells me I'm incorrect
 
cristo
Staff Emeritus
Science Advisor
8,056
72
Well, you haven't multiplied them by -1/5. In order to get [tex]\frac{-5}{210h+1764}[/tex] into the required form, you need to multiply top and bottom by -1/5 (to get unity in the numerator).
 
143
0
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
3,344
4
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.
 
143
0
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.
 
143
0
If not c=17, I don't understand what else it could be!
 

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