# Derivative using a limit definition

• Apr 6th 2012, 03:49 AM
Nforce
Derivative using a limit definition
It's pretty straight forward procedure.

We have a function:

$\displaystyle f(x) = 2x^2 + 3x$

To get the derivative, we use the limit definition.

$\displaystyle \lim_{h\to\0}\frac{f(x + h) - f(x)}{h}$

$\displaystyle \lim_{h\to\0}\frac{2(x+h)^2 + 3(x+h) - 2x^2 +3x}{h}$

$\displaystyle \lim_{h\to\0}\frac{2x^2 + 2xh + h^2 + 3x + 3h - 2x^2 +3x}{h}$

$\displaystyle \lim_{h\to\0}\frac{4xh + 2h^2 +3h + 6x}{h}$

But how can I get rid of 6x in the numerator? Because that's the problem, I can't get the proper derivative, if the 6x is in the numerator.
• Apr 6th 2012, 03:56 AM
princeps
Re: Derivative using a limit definition
Quote:

Originally Posted by Nforce
It's pretty straight forward procedure.

We have a function:

$\displaystyle f(x) = 2x^2 + 3x$

To get the derivative, we use the limit definition.

$\displaystyle \lim_{h\to\0}\frac{f(x + h) - f(x)}{h}$

$\displaystyle \lim_{h\to\0}\frac{2(x+h)^2 + 3(x+h) - 2x^2 +3x}{h}$

$\displaystyle \lim_{h\to\0}\frac{2x^2 + 2xh + h^2 + 3x + 3h - 2x^2 +3x}{h}$

$\displaystyle \lim_{h\to\0}\frac{4xh + 2h^2 +3h + 6x}{h}$

But how can I get rid of 6x in the numerator? Because that's the problem, I can't get the proper derivative, if the 6x is in the numerator.

second line :

$\displaystyle +3x$ should be $\displaystyle -3x$ because $\displaystyle -f(x)=-2x^2-3x$
• Apr 6th 2012, 03:58 AM
a tutor
Re: Derivative using a limit definition
Beaten to it by princeps.(Crying)
• Apr 6th 2012, 04:37 AM
Nforce
Re: Derivative using a limit definition
Oh, that thing I didn't saw. Thank you.
• Apr 6th 2012, 06:52 AM
Sylvia104
Re: Derivative using a limit definition
In fact you made a few mistakes.

Quote:

Originally Posted by Nforce
$\displaystyle \lim_{h\to\0}\frac{f(x + h) - f(x)}{h}$

$\displaystyle \lim_{h\to\0}\frac{2(x+h)^2 + 3(x+h) - 2x^2 \color{red}-\color{black}3x}{h}$

$\displaystyle \lim_{h\to\0}\frac{2x^2 + \color{red}4\color{black}xh + \color{red}2\color{black}h^2 + 3x + 3h - 2x^2 \color{red}-\color{black}3x}{h}$

• Apr 6th 2012, 09:17 AM
sbhatnagar
Re: Derivative using a limit definition
\displaystyle \begin{aligned} y =x^n \quad \text{then} \quad \frac{dy}{dx}&= \lim_{h \to 0} \frac{(x+h)^n-x^n}{h} \\ &=\lim_{h \to 0}\left[\frac{1}{h}\sum_{r=0}^{n}\binom{n}{r}x^{n-r}h^r-\frac{x^n}{h}\right] \\ &= \lim_{h \to 0}\left[ nx^{n-1}+\binom{n}{2}x^{n-2}h + \cdots +h^{n-1}\right] \\ &= nx^{n-1}\end{aligned}

This fact can now be easily applied to the original problem.