Derivative using first principles

**skyblue** · August 29th 2009, 09:41 AM

So I just finished my first week of calc and our weekend homework is to solve this one problem. We have to find the derivative of $y= x/ sqrt(3-x)$ using first principles.
I have no idea what I'm doing. It seems that every time I try to solve it I end up with a terribly long equation that doesn't seem to simplify. Any useful suggestions are welcome. Just please take into account I have just started learning calc.

**adkinsjr** · August 29th 2009, 10:11 AM

Originally Posted by skyblue

So I just finished my first week of calc and our weekend homework is to solve this one problem. We have to find the derivative of $y= x/ sqrt(3-x)$ using first principles.
I have no idea what I'm doing. It seems that every time I try to solve it I end up with a terribly long equation that doesn't seem to simplify. Any useful suggestions are welcome. Just please take into account I have just started learning calc.

You want to find:

$\frac{dy}{dx}=\lim_{h->0}\frac{f(x+h)-f(x)}{h}$

where

$f(x)=\frac{x}{\sqrt{3-x}}$

right?

**skyblue** · August 29th 2009, 10:19 AM

Originally Posted by adkinsjr

You want to find:

$\frac{dy}{dx}=\lim_{h->0}\frac{f(x+h)-f(x)}{h}$

where

$f(x)=\frac{x}{\sqrt{3-x}}$

right?

Yes exactly. Sorry I didn't know how to do the symbols...

**adkinsjr** · August 29th 2009, 11:12 AM

Ok, this one is kind of tricky because there is alot of algebra.

$\lim_{h->0}(\frac{x+h}{h\sqrt{3-(x+h)}}-\frac{x}{h\sqrt{3-x}})$

= $\lim_{h->0}\frac{h(x+h)\sqrt{3-x}-xh\sqrt{3-(x+h)}}{h\sqrt{3-(x+h)}\sqrt{3-x}}$

= $\lim_{h->0}\frac{xh\sqrt{3-x}+h^2\sqrt{3-x}-xh\sqrt{3-(x+h)}}{h\sqrt{3-(x+h)}\sqrt{3-x}}$

Hold on, I'll finish this up for you in a min, I need to restart my computer. sorry, I was having trouble typing the fractions. There's alot of algebra here lol. Do you get this far when you try to work it out?

**Soroban** · August 29th 2009, 01:20 PM

Hello, skyblue!

This is the hardest First Principle problem I've ever seen!
In case adkinsjr doesn't get back soon, I'll crank it out . . .

Find the derivative of $f(x)\:=\: \frac{x}{\sqrt{3-x}}$ using first principles.

I recommend breaking up the definition into four steps: . $f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}$

. . [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 reduce.

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

Are you ready? . Here we go . . .

[1] $f(x+h) \;=\;\frac{x+h}{\sqrt{3-(x+h)}} \;=\;\frac{x+h}{\sqrt{3-x-h}}$

[2] $f(x+h) - f(x) \;=\;\frac{x+h}{\sqrt{3-x-h}} - \frac{x}{\sqrt{3-x}} \;=\;\frac{(x+h)\sqrt{3-x} - x\sqrt{3-x-h}}{\sqrt{3-x}\cdot\sqrt{3-x-h}}$

. . Multiply top and bottom by the conjugate of the numerator:

. . . . $\frac{(x+h)\sqrt{3-x} - x\sqrt{3-x-h}}{\sqrt{3-x}\cdot\sqrt{3-x-h}}\cdot\frac{(x+h)\sqrt{3-x} + x\sqrt{3-x-h}}{(x+h)\sqrt{3-x} + x\sqrt{3-x-h}} \;=$ . $\frac{(x+h)^2(3-x) - x^2(3-x-h)}{\sqrt{3-x}\cdot\sqrt{3-x-h}\cdot[(x+h)\sqrt{3=x} + x\sqrt{3-x-h}]}$

. . which simplifies to: . $\frac{6xh - x^2h - xh^2 + 3h^2}{\text{denominator}} \;=\; \frac{h\,(6x - x^2 - xh + 3h)}{\text{denominator}}$

[3] $\frac{f(x+h) - f(x)}{h} \;=\;\frac{1}{{\color{red}\rlap{/}}h}\cdot\frac{{\color{red}\rlap{/}}h\,(6x - x^2 - xh + 3h)}{\text{denominator}} \;=\;\frac{6x - x^2 - xh + 3h}{\text{denominator}}$

[4] $f'(x) \;=\;\lim_{h\to0}\left[ \frac{6x-x^2 - xh + 3h}{\sqrt{3-x}\cdot\sqrt{3-x-h}\cdot[(x+h)\sqrt{3-x} + x\sqrt{3-x-h}]}\right]$

. . . . . . $= \;\frac{6x - x^2 - 0 + 0}{\sqrt{3-x}\cdot\sqrt{3-x-0}\cdot\left[(x+0)\sqrt{3-x} + x\sqrt{3-x-0}\right]}$

. . . . . . $= \;\frac{6x-x^2}{\sqrt{3-x}\cdot\sqrt{3-x}\cdot(x\sqrt{3-x} + x\sqrt{3-x})}$

. . . . . . $= \;\frac{6x - x^2}{(3-x)2x\sqrt{3-x}} \;\;=\;\;\frac{6x-x^2}{2x(3-x)^{\frac{3}{2}}} \;\;=\;\;\frac{{\color{red}\rlap{/}}x(6-x)}{2{\color{red}\rlap{/}}x(3-x)^{\frac{3}{2}}} \;\;=\;\; \boxed{\frac{6-x}{2(3-x)^{\frac{3}{2}}}}$

I need a nap!
.

**Alterah** · August 29th 2009, 06:42 PM

Wow...I was watching that and it is a tough one to assign. Significantly easier with the quotient rule...but that wasn't part of the directions...

**adkinsjr** · August 29th 2009, 08:16 PM

Thanks folks, I started to have trouble with that one too. Students are usually asked to find the derivatives of functions like $y=x^2$ or $y=\sqrt{1-x}$ using this method. More complex derivatives are just found using formulae for products and quotients, which can be applied with zero intelligence.

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