I am having some difficulty finding the derivative of this equation:
http://i49.tinypic.com/35jdoxt.png
Can anyone assist me in solving this?
I am having some difficulty finding the derivative of this equation:
http://i49.tinypic.com/35jdoxt.png
Can anyone assist me in solving this?
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 $\displaystyle \displaystyle \begin{align*} g'(x) = \frac{\sqrt{10}}{2\sqrt{x}} \end{align*}$.