# A derivative problem?

• Oct 10th 2008, 05:28 PM
_nysa
A derivative problem?
Here's the problem I'm having.

f(x)=ax^n, where n and a are constants and f'(2)=3 and f'(4)= 24

I'm totally confused as to what I should be doing of even what the first steps should be. Any help is greatly appreciated.
• Oct 10th 2008, 05:35 PM
Jhevon
Quote:

Originally Posted by _nysa
Here's the problem I'm having.

f(x)=ax^n, where n and a are constants and f'(2)=3 and f'(4)= 24

I'm totally confused as to what I should be doing of even what the first steps should be. Any help is greatly appreciated.

they want you to find the constants a and n

find the derivative, then, plug in the values as prescribed. you will obtain two equations in a and n that you can solve simultaneously
• Oct 10th 2008, 05:39 PM
_nysa
Quote:

they want you to find the constants a and n

find the derivative, then, plug in the values as prescribed. you will obtain two equations in a and n that you can solve simultaneously
Maybe I'm being slow here, but when I do a derivative I get:

a(nx^n-1)

And this is the part where I get confused, because I don't know what values go where.
• Oct 10th 2008, 05:45 PM
Chris L T521
A similar problem was done here.

--Chris
• Oct 10th 2008, 06:07 PM
_nysa
Thanks Chris, but I'm stuck again.

I followed the example that you linked too, and this is what I got to.

1/8 = (1/8)^-1(1/8)^n

And then I don't know what to do anymore.
• Oct 10th 2008, 06:17 PM
Chris L T521
Quote:

Originally Posted by _nysa
Thanks Chris, but I'm stuck again.

I followed the example that you linked too, and this is what I got to.

1/8 = (1/8)^-1(1/8)^n

And then I don't know what to do anymore.

Well, $\left(\tfrac{1}{8}\right)^{-1}=8$

So we have $\tfrac{1}{8}=8\cdot\left(\tfrac{1}{8}\right)^n$

This implies $\tfrac{1}{64}=\left(\tfrac{1}{8}\right)^n$

Can you finish this?

[hint: $64=8^2$]

--Chris
• Oct 10th 2008, 06:29 PM
_nysa
Yes I can, thanks so much!