derivative by first principles

**ArTiCK** · March 29th 2008, 07:53 PM

Hi all,
I was wondering how do you find the derivative of the following function by first principles:

(1 + 2x) ^(1/2)

When i try to find the derivate by first principles the x cancels out. Well thank you in advance.

ArTiCk

**o_O** · March 29th 2008, 08:53 PM

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

$f'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 2(x + h)} - \sqrt{1+2x}}{h}$

$f'(x) = \lim_{h \to 0} \frac{\sqrt{1 + 2x + 2h} - \sqrt{1+2x}}{h}$

Rationalize the numerator:

$f'(x) = \lim_{h \to 0} \left( \frac{\sqrt{1 + 2x + 2h } - \sqrt{1+2x}}{h} \cdot \frac{\sqrt{1+ 2x + 2h} + \sqrt{1+2x}}{\sqrt{1+ 2x + 2h} + \sqrt{1+2x}}\right)$

$f'(x) = \lim_{h \to 0} \frac{1 + 2x + 2h - (1 + 2x)}{h \left(\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}\right)}$

And simplify. Yes the x's cancel out in the numerator but they're still there in the denominator.

**Soroban** · March 29th 2008, 09:17 PM

Hello, ArTiCK!

Find the derivative by first principles: . $f(x) \:=\:\sqrt{1 + 2x}$

We want: . $f'(x) \;=\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

(1) Find $f(x+h)$

. . $f(x+h) \:=\:\sqrt{1 + 2(x+h)}$

(2) Subtract $f(x)$

. . $f(x+h) - f(x) \:=\:\sqrt{1 + 2(x+h)} - \sqrt{1+2x}$

Rationalize: . $\frac{\sqrt{1+2(x+h)} - \sqrt{1+2x}}{1}\cdot\frac{\sqrt{1+2(x+h)} + \sqrt{1+2x}}{\sqrt{1+2(x+h)} + \sqrt{1+2x}}$

. . $= \;\frac{[1 + 2(x+h)] - [1 + 2x]}{\sqrt{1+2(x+h)} + \sqrt{1 + 2x}} \;=$ . $\frac{1 + 2x + 2h - 1 - 2x}{\sqrt{1+2(x+h)} + \sqrt{1+2x}} \;=\;\frac{2h}{\sqrt{1+2(x+h)} + \sqrt{1+2x}}$

(3) Divide by $h.$

. . $\frac{f(x+h)-f(x)}{h} \;=\;\frac{2h}{h[\sqrt{1 + 2(h+x)} + \sqrt{1+2x}} \;=\;\frac{2}{\sqrt{1+2(x+h)} + \sqrt{1+2x}}$

(4) Take the limit.

. . $\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{x\to0}\frac{2}{\sqrt{1+2(x+h)} + \sqrt{1+2x}} \;=\;\frac{2}{\sqrt{1+2x}\cdot\sqrt{1+2x}}$

. . . . . $= \;\frac{2}{2\sqrt{1+2x}} \;=\;\frac{1}{\sqrt{1+2x}}$

Therefore: . $\frac{d}{dx}\left(\sqrt{1+2x}\right) \;=\;\frac{1}{\sqrt{1+2x}}$

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