# Thread: Finding the Derivative of the Given Function

1. ## Finding the Derivative of the Given Function

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

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

Can anyone assist me in solving this?

2. ## Re: Finding the Derivative of the Given Function

First, I would rewrite the function as \displaystyle \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 \displaystyle \begin{align*} \sqrt{x} \end{align*}.

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

3. ## Re: Finding the Derivative of the Given Function

Thanks so much!

4. ## Re: Finding the Derivative of the Given Function

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...

5. ## Re: Finding the Derivative of the Given Function

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