Derivative of $\sqrt{\frac{9+x}{x}}$ using first principle

I am trying to find the derivative of $\sqrt{\dfrac{9+x}{x}}$ using first principle to differentiate.

For sake of simplifying the formulas, I will omit the /h part to the first principles formula, unless someone is able to show me how to integrate it with these equations.

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

What I tried was rationalizing/simplifying first:

$$\dfrac{\sqrt{9+x}}{x} \cdot \dfrac{\sqrt{9+x}}{\sqrt{9+x}}=\dfrac{9+x}{x\sqrt{9+x}}$$

Which when subbed into original first principle formula you get:

$$\dfrac{(9+x+h)}{(x+h)\sqrt{9+x}} - \dfrac{(9+x)}{x\sqrt{9+x}}$$

I then tried to find a common denominator and this was the result:

$$\dfrac{(9x+x^2+h)x\sqrt{9+x}}{x^2+18+2x+xh}- \dfrac{(9x+9h+x^2+hx)\sqrt{9+x}}{x^2+18+2x+xh}$$

After collecting like terms I get:

$$\dfrac{\dfrac{1}{x^2+18+2x+xh} - \dfrac{9h}{x^2+18+2x+xh}}{h}$$

I know it is close because the numerator is close to what the final product looks like: $$\dfrac{-9}{2 x^2 \sqrt{\dfrac{x+9}{x}}}$$

• You're a very brave hooman. – Pedro Tamaroff Feb 15 '14 at 23:14
• Is your expression $\sqrt{\dfrac{9+x}x}$ or $\dfrac{\sqrt{9+x}}x$? – Alraxite Feb 15 '14 at 23:20
• It is the first one, apologies it is my first time using mathjax – Reese Feb 15 '14 at 23:21

For $f(x)=\sqrt {\dfrac{9+x}{x}}$, \begin{align} &\dfrac{f(x+h)-f(x)}{h} \\=&\dfrac{\sqrt{\dfrac{9+x+h}{x+h}}-\sqrt{\dfrac{9+x}{x}}}{h} \\\\\\=&\dfrac{{\dfrac{9+x+h}{x+h}}-{\dfrac{9+x}{x}}}{h\left(\sqrt{\dfrac{9+x+h}{x+h}}+\sqrt{\dfrac{9+x}{x}}\right)}\tag{Multiplying by the conjugate} \end{align}
The numerator simplifies as $\dfrac{9x+x^2+hx-(x^2+9h+hx+9x)}{x(x+h)}=\dfrac{-9h}{x(x+h)}$.
So, $\dfrac{f(x+h)-f(x)}{h}=\dfrac{ \dfrac{-9}{x(x+h)} }{ \left(\sqrt{\dfrac{9+x+h}{x+h}}+\sqrt{\dfrac{9+x}{x}}\right) }$.
Now as $h\to0$, $\dfrac{f(x+h)-f(x)}{h}\to\dfrac{ \dfrac{-9}{x^2} }{2\sqrt{\dfrac{9+x}{x}}}=\dfrac{-9}2 \dfrac{\sqrt{x}}{x^2\sqrt{9+x}}$ which is our sought derivative.