# Finding a derivative using first principles

• Jun 10th 2010, 08:55 PM
Glitch
Finding a derivative using first principles
I'm working on this question:

if f(x) = x^3, then the derivative is 3x^2

I have to prove this using the definition of the derivative. I got as far as this:

$((x^3)h + 2(h^2)(x^2)+(h^3)x-x^3)/(h)$

I'm unsure of how to simplify it in order to find the limit as h -> 0.
• Jun 10th 2010, 09:18 PM
StudyBug10
lim h --> 0 of x^3 = [(x+h)^3 - x^3]/h
= [(x+h)(x+h)(x+h) - x^3]/h
= [x^3 + x^2 h + 2 x^2 h + 2 x h^2 + x h^2 + h^3 - x^3]/h
cancel the x^3 and the h's with the one in the bottom
= x^2 + 2 x^2 + 2 x h + x h + h^2
plug in 0 for h
= x^2 + 2 x^2
combine like terms
= 3 x^2

I hope that helps :D
• Jun 10th 2010, 09:31 PM
Glitch
Ahh, thanks! The problem was with my simplification of (x+h)^2
• Jun 10th 2010, 10:33 PM
Glitch
I have a similar question:

Find the derivative of $1/x$ using first principles. Once again, I'm having trouble simplifying. :/

I keep getting $0/x^2$ when it should be $-1/x^2$
• Jun 10th 2010, 11:16 PM
mr fantastic
Quote:

Originally Posted by Glitch
I have a similar question:

Find the derivative of $1/x$ using first principles. Once again, I'm having trouble simplifying. :/

I keep getting $0/x^2$ when it should be $-1/x^2$

• Jun 10th 2010, 11:30 PM
Glitch
Working:

$\frac{\frac{1}{x+h} - \frac{1}{x}}{h}$

$\frac{h}{x+h} - \frac{h}{x}$

$\frac{hx - h(x+h)}{x(x+h)}$

$\frac{hx - h(x+h)}{x^2 + hx}$

As h -> 0,

$0/x^2$

Which is incorrect.
• Jun 10th 2010, 11:34 PM
mr fantastic
Quote:

Originally Posted by Glitch
Working:

$\frac{\frac{1}{x+h} - \frac{1}{x}}{h}$

$\frac{h}{x+h} - \frac{h}{x}$ Mr F says: This is wrong. All that follows from it will therefore be wrong. It should be ${\color{red}\frac{1}{(x+h)h} - \frac{1}{xh}}$.

$\frac{hx - h(x+h)}{x(x+h)}$

$\frac{hx - h(x+h)}{x^2 + hx}$

As h -> 0,

$0/x^2$

Which is incorrect.

..