Derivative using a limit definition

**Nforce** · April 6th 2012, 03:49 AM

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.

**princeps** · April 6th 2012, 03:56 AM

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

**a tutor** · April 6th 2012, 03:58 AM

Beaten to it by princeps.

**Nforce** · April 6th 2012, 04:37 AM

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

**Sylvia104** · April 6th 2012, 06:52 AM

In fact you made a few mistakes.

Originally Posted by Nforce

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

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

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

**sbhatnagar** · April 6th 2012, 09:17 AM

$\begin{aligned} y =x^n \quad \text{then} \quad \frac{dy}{dx}&= \lim_{h \to 0} \frac{(x+h)^n-x^n}{h} \\ &=\lim_{h \to 0}\left[\frac{1}{h}\sum_{r=0}^{n}\binom{n}{r}x^{n-r}h^r-\frac{x^n}{h}\right] \\ &= \lim_{h \to 0}\left[ nx^{n-1}+\binom{n}{2}x^{n-2}h + \cdots +h^{n-1}\right] \\ &= nx^{n-1}\end{aligned}$

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

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