# Thread: A derivative problem?

1. ## 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.

2. 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

3. 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.

4. A similar problem was done here.

--Chris

5. 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.

6. 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

7. Yes I can, thanks so much!