# Solving Derivatives

• Sep 21st 2013, 11:20 AM
vaironxxrd
Solving Derivatives
Hello Everyone,

In preparation for an upcoming exam I've selected a derivative problem and attempted to solve it...

I have to use $f^'$ $(x) = \frac{f(x+h)-f(x)}{h}$ to solve these problems.
My attempts:

a) $f(x) = 3x^3$ Using the book theorems I could quickly find the answer was $9x^2$ Now I have to prove this...

$f^'$ $(x) = \frac{3(x+h)^3-3x^3}{h}$ =

$\frac{3h^3+9h^2x+9hx^2+3x^3}{h}$ = (cancel h)

$9x^2+3x^3$

This solution of course doesn't match $9x^2$
• Sep 21st 2013, 12:24 PM
Plato
Re: Solving Derivatives
Quote:

Originally Posted by vaironxxrd
Hello Everyone,

In preparation for an upcoming exam I've selected a derivative problem and attempted to solve it...

I have to use $f^'$ $(x) = \frac{f(x+h)-f(x)}{h}$ to solve these problems.
My attempts:

a) $f(x) = 3x^3$ Using the book theorems I could quickly find the answer was $9x^2$ Now I have to prove this...

$f^'$ $(x) = \frac{3(x+h)^3-3x^3}{h}$ =

$\frac{3h^3+9h^2x+9hx^2+3x^3}{h}$ = (cancel h)

$9x^2+3x^3$

This solution of course doesn't match $9x^2$

It should be:
$\frac{3h^3+9h^2x+9hx^2}{h}$ = (cancel h)

$9x^2$
• Sep 21st 2013, 12:39 PM
vaironxxrd
Re: Solving Derivatives
Quote:

Originally Posted by Plato
It should be:
$\frac{3h^3+9h^2x+9hx^2}{h}$ = (cancel h)

$9x^2$

I see I totally forgot the -3x^3.
I see these (x+h)^n term are getting a little bit longer on some problems.