Author Topic: first principles (Read 734 times) Tweet Share

boysenberry · « **on:** November 14, 2009, 01:07:02 pm »

Use first principles to find $\frac{dy}{dx}$ if $y = \frac{x^2}{3}$ .

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

Also, excuse my ignorance.

dekoyl · « **Reply #1 on:** November 14, 2009, 01:17:14 pm »

Eh I haven't done this since last year so I'm not 100% sure
$\frac{dy}{dx} = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}$

$=\frac{\frac{(x+h)^2}{3} - \frac{x^2}{3}}{h}$

$=\frac{\frac{2xh}{3}}{h}$

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

Woops. Dejan91 is right - I'm not.

dejan91 · « **Reply #2 on:** November 14, 2009, 01:20:54 pm »

Let $f(x) = \frac{x^{2}}{3}$

$\frac{dy}{dx} = f'(x)= \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} =\lim_{h\rightarrow 0} \frac{x^2 + 2hx + h^2 - x^2}{3h}}} =\lim_{h\rightarrow 0} \frac{h(2x + h )}{3h} =\lim_{h\rightarrow 0} \frac{(2x + h )}{3} = \frac{2x}{3}$

Note that you have to include $\lim_{h\rightarrow 0}$ in all of them but the last expression.

boysenberry · « **Reply #3 on:** November 14, 2009, 01:35:40 pm »

Quote from: dejan91 on November 14, 2009, 01:20:54 pm

Let $f(x) = \frac{x^{2}}{3}$

$\frac{dy}{dx} = f'(x)= \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} =\lim_{h\rightarrow 0} \frac{x^2 + 2hx + h^2 - x^2}{3h}}} =\lim_{h\rightarrow 0} \frac{h(2x + h )}{3h} =\lim_{h\rightarrow 0} \frac{(2x + h )}{3} = \frac{2x}{3}$

Note that you have to include $\lim_{h\rightarrow 0}$ in all of them but the last expression.

Thank you for you help!

crappy · « **Reply #4 on:** November 14, 2009, 02:30:19 pm »

damn, after all these years of maths, I finally understand why first principles is so important, its bloody ingenious!

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