# Math Help - Derivative using a limit definition

1. ## Derivative using a limit definition

It's pretty straight forward procedure.

We have a function:

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

To get the derivative, we use the limit definition.

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

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

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

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

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.

2. ## Re: Derivative using a limit definition

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

We have a function:

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

To get the derivative, we use the limit definition.

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

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

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

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

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 :

$+3x$ should be $-3x$ because $-f(x)=-2x^2-3x$

3. ## Re: Derivative using a limit definition

Beaten to it by princeps.

4. ## Re: Derivative using a limit definition

Oh, that thing I didn't saw. Thank you.

5. ## Re: Derivative using a limit definition

In fact you made a few mistakes.

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

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

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

6. ## Re: Derivative using a limit definition

\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.