Question

the derivative of g(x)= f(x)^3
using the definition

Answer 1

Are you familiar with the chain rule?

Answer 2

can't use the chain rule or I would
it has to be done with the definition

Answer 3

Use the chain rule

Answer 4

what is the derivative of x^3?

Answer 5

Just to make sure, this is still the derivative of
\[\left[f(x)\right]^3, \text{ right?}\]
Or, is it
\[f\left(x^3\right)?\]

Answer 6

the way it is set up is \[f(x)^3 \]

Answer 7

Okay, then that's the first one I listed.
\[g'(x) = \lim_{h\to 0}\frac{\left[f(x+h)\right]^3-\left[f(x)\right]^3}{h}\]
Just a moment, let me work this out on paper (if I can).

Answer 8

lol ok sounds good

Answer 9

is the question asking what the derivative of g(x) is?

Answer 10

the derivative of \[g(x)=f(x)^3\]

Answer 11

so g'(x) = f(x)^3?

Answer 12

Okay, I think I've got it. However, it's not as intuitive as it should be. I'm going to ignore the limit for a while and just rewrite/rearrange the numerator for a bit (hope it doesn't get cut off):
\[f(x+h)^3-f(x)^3\\
f(x+h)^3-f(x)^3 + \left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]-\left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]\\
f(x+h)^3-f(x)^3-\left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right] + \left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]\\
\left[f(x+h)^3-f(x)^3-3f(x)f(x+h)^2+3f(x)^2f(x+h)\right] + \left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]\\
\left[f(x+h)^3-3f(x)f(x+h)^2+3f(x)^2f(x+h)-f(x)^3\right] + \left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]\\
\left[f(x+h)-f(x)\right]^3 + \left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]\]
Is that all clear?

Answer 13

Darn, it did get cut off... The cut-off portion is
\[...-3f(x)^2f(x+h)\]
I only rewrite the first big sum, and the cut-off section reappears perfectly in the last line.

Answer 14

I know it seems like I pulled that [3f(x)²f(x+h) - 3f(x)f(x+h)^2] term out of nowhere, but it's all for a reason.

Answer 15

ok thats fine... I just haven't been able to figure them out

Answer 16

So does it all make at least some sense now?

Answer 17

umm kind of

Answer 18

Okay, so the last line that I typed out is now the numerator of the limit:
\[\lim_{h\to 0}\frac{\left[f(x+h)-f(x)\right]^3 + \left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]}{h}\\
\lim_{h\to 0}\frac{\left[f(x+h)-f(x)\right]^3}{h} + \lim_{h\to 0}\frac{\left[3f(x)f(x+h)^2-3f(x)^2f(x+h)\right]}{h}\]
First, let's deal with the first limit:
\[\lim_{h\to 0}\frac{\left[f(x+h)-f(x)\right]^3}{h}\\
\lim_{h\to 0}\frac{\left[f(x+h)-f(x)\right]\left[f(x+h)-f(x)\right]^2}{h}\\
\lim_{h\to 0}\left[f(x+h)-f(x)\right]\frac{\left[f(x+h)-f(x)\right]^2}{h}\\
\lim_{h\to 0}\left[f(x+h)-f(x)\right]\cdot\lim_{h\to 0}\frac{\left[f(x+h)-f(x)\right]^2}{h}\\
\left[f(x)-f(x)\right]\cdot\lim_{h\to 0}\frac{\left[f(x+h)-f(x)\right]^2}{h}\\
0\cdot\lim_{h\to 0}\frac{\left[f(x+h)-f(x)\right]^2}{h}\\
0\]

Answer 19

Now the second limit:
\[\lim_{h\to 0}\frac{3f(x)f(x+h)^2-3f(x)^2f(x+h)}{h}\\
\lim_{h\to 0}\frac{3f(x)f(x+h)\left[f(x+h)-f(x)\right]}{h}\\
\lim_{h\to 0}\left[3f(x)f(x+h)\right]\cdot\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\
3f(x)f(x)\cdot f'(x)\\
3f(x)^2\cdot f'(x)\]
Finally done!

Answer 20

ok cool thank you