Derivative (1st Principles)

**millerst** · February 17th 2009, 02:52 PM

Find the derivative of the function using the definition of derivative.

1) x + sqrt(x)

2) 1 / sqrt(x)

---------------
1) ((x+h) + sqrt(x+h) - [x + sqrt(x)])/h
I'm stuck here, not sure what to do?

2) [(1/sqrt(x+h)) - (1/sqrt(x))]/h
Not sure what to do here either?

**Soroban** · February 17th 2009, 03:52 PM

Hello, millerst!

This is a test of your algebraic ability and stamina.
I'll do the first one (it's harder).

Find the derivative of the function using the definition of derivative.

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

Definition of derivative: . $f\:\!'(x) \;=\;\lim_{h\to0} \frac{f(x+h) - f(x)}{h}$

My recommendation: divide the procedure into four steps:

. . [1] Find $f(x+h)$ . . . replace $x$ with $x+h$ ... and simplify.

. . [2] Subtract $f(x)$ . . . subtract the original function ... and simplify.

. . [3] Divide by $h$ . . .factor and simplify.

. . [4] Take the limit as $h\to0.$

Here we go . . .

[1] Find $f(x+h)\!:\quad f(x+h) \;=\;(x+h) + \sqrt{x+h}$

[2] Subtract $f(x)\!:\quad \bigg[(x+h) + \sqrt{x+h}\bigg] - \bigg[x + \sqrt{x}\bigg] \;=\;h + \sqrt{x+h} - \sqrt{x}$

. . . $\text{Multiply by: }\,\frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h}+\sqrt{x}}:\quad \frac{h+\sqrt{x+h}-\sqrt{x}}{1}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{ x+h}+\sqrt{x}}$

. . . . . $=\;\frac{h(\sqrt{x+h} + \sqrt{x}) + (\sqrt{x+h} - \sqrt{x})(\sqrt{x+h}+\sqrt{x})} {\sqrt{x+h}+x\sqrt{x}}$

. . . . . $= \;\frac{h(\sqrt{x+h} + \sqrt{x}) + (x + h - x)}{\sqrt{x+h}+\sqrt{x}}$

. . . . . $= \;\frac{h(\sqrt{x+h} + \sqrt{x}) + h}{\sqrt{x+h} + \sqrt{x}} \;=\; \frac{h\,(\sqrt{x+h}+\sqrt{x} + 1)}{\sqrt{x+h}+\sqrt{x}}$

[3] Divide by $h\!:\quad \frac{1}{{\color{red}\rlap{/}}h}\cdot\frac{{\color{red}\rlap{/}}h\left(\sqrt{x+h} + \sqrt{x} + 1\right)}{\sqrt{x+h} + \sqrt{x}} \:=\:\frac{\sqrt{x+h}+\sqrt{x}+1}{\sqrt{x+h}+\sqrt {x}}$

[4] Take the limit:

. . $\lim_{h\to0}\frac{\sqrt{x+h} + \sqrt{x} + 1}{\sqrt{x+h} + \sqrt{x}} \;=\;\frac{\sqrt{x} + \sqrt{x} + 1}{\sqrt{x} + \sqrt{x}} \;= \;\frac{2\sqrt{x} + 1}{2\sqrt{x}} \;=\;\frac{2\sqrt{x}}{2\sqrt{x}} + \frac{1}{2\sqrt{x}}$

Therefore: . $f\:\!'(x) \;=\;1 + \frac{1}{2\sqrt{x}}$

I need a nap . . .
.

**millerst** · February 17th 2009, 04:10 PM

Hi, thanks for the reply. I'm confused at this step:

$<br /> <br /> = \;\frac{h(\sqrt{x+h} + \sqrt{x}) + h}{\sqrt{x+h} + \sqrt{x}} \;=\; \frac{h\,(\sqrt{x+h}+\sqrt{x} + 1)}{\sqrt{x+h}+\sqrt{x}}<br />$

How did the + h, go into the bracket and become + 1?

**Soroban** · February 17th 2009, 04:18 PM

Hello, millerst

I'm confused at this step:

$\frac{h(\sqrt{x+h} + \sqrt{x}) + h}{\sqrt{x+h} + \sqrt{x}} \;=\; \frac{h\,(\sqrt{x+h}+\sqrt{x} + 1)}{\sqrt{x+h}+\sqrt{x}}<br />$

How did the + h, go into the bracket and become + 1?

We factor out the $h.$

**millerst** · February 17th 2009, 04:24 PM

Wow, I can't believe I didn't see that. Thanks so much.

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