# Finding the Derivative of the Given Function

• Feb 2nd 2013, 08:24 PM
michellederz
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?
• Feb 2nd 2013, 08:56 PM
Prove It
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*}.
• Feb 2nd 2013, 09:02 PM
michellederz
Re: Finding the Derivative of the Given Function
Thanks so much!
• Feb 2nd 2013, 09:04 PM
Prove It
Re: Finding the Derivative of the Given Function
Quote:

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...
• Feb 2nd 2013, 09:05 PM
michellederz
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..