# Derivative from first principles

• Nov 13th 2009, 04:44 PM
Joker37
Derivative from first principles
Use first principles to find $\frac{dy}{dx}$ if $\frac{x^2}{3}$.

I've never used the first principles to differentiate a fraction. Could someone please help me with the above and supply correct working out. All help will be appreciated.
• Nov 13th 2009, 04:48 PM
Quote:

Originally Posted by Joker37
Use first principles to find $\frac{dy}{dx}$ if $\frac{x^2}{3}$.

I've never used the first principles to differentiate a fraction. Could someone please help me with the above and supply correct working out. All help will be appreciated.

HI

$y=\frac{1}{3}x^2$

$
\frac{dy}{dx}=2(\frac{1}{3})x^{2-1}
$
• Nov 13th 2009, 04:53 PM
Joker37
Quote:

Originally Posted by mathaddict
HI

$y=\frac{1}{3}x^2$

$
\frac{dy}{dx}=2(\frac{1}{3})x^{2-1}
$

No, no...you have to use first principles to differentiate it.
• Nov 13th 2009, 05:28 PM
Quote:

Originally Posted by Joker37
No, no...you have to use first principles to differentiate it.

oh ok

$\frac{d}{dx}(\frac{x^2}{3})=\frac{1}{3}\frac{d}{dx }(x^2)$

$
\frac{1}{3}\lim_{\delta{x}\rightarrow 0}[\frac{(x+\delta{x})^2-x^2}{\delta{x}}]
$

$\frac{1}{3}\lim_{\delta \rightarrow 0}[\frac{x^2+2x\delta{x}+(\delta{x})^2-x^2}{\delta{x}}]$

$\frac{1}{3}\lim_{\delta{x}\rightarrow 0}(2x+\delta{x})$

$
=\frac{2x}{3}
$