Derivative from first principles

**Joker37** · November 13th 2009, 04:44 PM

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.

**mathaddict** · November 13th 2009, 04:48 PM

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} $

**Joker37** · November 13th 2009, 04:53 PM

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.

**mathaddict** · November 13th 2009, 05:28 PM

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} $

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