Finding the Derivative of the Given Function

**michellederz** · February 2nd 2013, 08:24 PM

I am having some difficulty finding the derivative of this equation:

http://i49.tinypic.com/35jdoxt.png

Can anyone assist me in solving this?

**Prove It** · February 2nd 2013, 08:56 PM

First, I would rewrite the function as $\displaystyle \begin{align*} g(x) = \sqrt{10} \, \sqrt{x} \end{align*}$ and remember that the derivative of a function times a constant is equal to that constant times the derivative. So really, you just need to find the derivative of $\displaystyle \begin{align*} \sqrt{x} \end{align*}$ .

$\displaystyle \begin{align*} f(x) &= \sqrt{x} \\ \\ f(x + h) &= \sqrt{x + h} \\ \\ f'(x) &= \lim_{h \to 0}\frac{f(x+ h) - f(x)}{h} \\ &= \lim_{h \to 0}\frac{\sqrt{x + h} - \sqrt{x}}{h} \\ &= \lim_{h \to 0}\frac{\left( \sqrt{x + h} - \sqrt{x} \right) \left( \sqrt{x + h} + \sqrt{x} \right) }{h \left( \sqrt{x + h} + \sqrt{x} \right) } \\ &= \lim_{h \to 0}\frac{ x + h - x }{h \left( \sqrt{x + h} + \sqrt{x} \right) } \\ &= \lim_{h \to 0}\frac{h}{h \left( \sqrt{x + h} + \sqrt{x} \right) } \\ &= \lim_{h \to 0}\frac{1}{\sqrt{x + h} + \sqrt{x}} \\ &= \frac{1}{\sqrt{x + 0} + \sqrt{x}} \\ &= \frac{1}{\sqrt{x} + \sqrt{x}} \\ &= \frac{1}{2\sqrt{x}} \end{align*}$

Therefore if $\displaystyle \begin{align*} g(x) = \sqrt{10} \, \sqrt{x} \end{align*}$ then $\displaystyle \begin{align*} g'(x) = \frac{\sqrt{10}}{2\sqrt{x}} \end{align*}$ .

**michellederz** · February 2nd 2013, 09:02 PM

Thanks so much!

**Prove It** · February 2nd 2013, 09:04 PM

Originally Posted by michellederz

Thank you so much but, apparently the answer can be simplified but I don't see where... Any suggestions on how to simplify the answer you gave?

What do you consider simpler? What is the expected "simplified" answer? To me, this is the simplest you can get...

**michellederz** · February 2nd 2013, 09:05 PM

I figured it out. My math book just had it in a different form for whatever reason..

It put the answer as:

5/(square root of 10x)

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