Differentiation Using first principle Q (1 Viewer)

TearsOfFire · Apr 11, 2009

How do you differentiate this Q:

3 sqrt x

using first principle.

I got up to here and I got stuck; Not sure if its right though

3(sqrt x + h) - 3 sqrt x / h

Iruka · Apr 11, 2009

You know how to rationalise the denomiator of a fraction involving surds?

To answer this q, you rationalise the numerator, so you multiply by (sqrt(x+h)+sqrt(x))/(sqrt(x+h)+sqrt(x)). You can then simplify the numerator, and cancel the h in the denominator.

Drongoski · Apr 11, 2009

For those who may like it spelt out more fully here's my attempt:

Let f(x) = 3sqr(x)

$\therefore \ \ \frac{\mathrm{d} }{\mathrm{d} x}f(x)\ \ =\ \ \lim_{h\rightarrow 0}\frac{f(x+h)\ -\ f(x)}{h}\\ \\ =\ \ \lim_{h\rightarrow 0}\frac{3\sqrt{x+h}-\ 3\sqrt{x}}{h}\ \ =\ \ \lim_{h\to 0}\frac{3\left \{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x}) \right \}}{h(\sqrt{x+h}+\sqrt{x})}\\ \\ =\ \ \lim_{h\to 0}\frac{3(x+h\ -\ x)}{h(\sqrt{x+h}+\sqrt{x})}\ \ =\ \ \lim_{h\rightarrow 0}\frac{3}{\sqrt{x+h}+\sqrt{x}}\\ \\ =\ \ \frac{3}{\sqrt{x+0}\ +\ \sqrt{x}}\ \ =\ \ \frac{3}{2\sqrt{x}}$

Differentiation Using first principle Q (1 Viewer)

TearsOfFire

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Iruka

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Drongoski

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