# Thread: Derivative using first principles

1. ## Derivative using first principles

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 $\displaystyle y= x/ sqrt(3-x)$ 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.

2. 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 $\displaystyle y= x/ sqrt(3-x)$ 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:

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

where

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

right?

You want to find:

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

where

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

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

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

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

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

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

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?

5. 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 $\displaystyle f(x)\:=\: \frac{x}{\sqrt{3-x}}$ using first principles.

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

. . [1] Find $\displaystyle f(x+h)$ . . . Replace $\displaystyle x$ with $\displaystyle x+h$ ... and simplify.

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

. . [3] Divide by $\displaystyle h$ . . . Factor and reduce.

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

Are you ready? . Here we go . . .

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

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

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

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

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

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

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

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

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

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

I need a nap!
.

6. 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...

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