\(\displaystyle f(x) = \dfrac{9}{\sqrt{x}} \)
\(\displaystyle f(x) = \dfrac{9}{x^{1/2}} \)
\(\displaystyle f(x) = 9x^{-1/2} \)
\(\displaystyle f'(x) = \lim h \to 0[\dfrac{9(x + h)^{-1/2} - (9x^{1/2)})}{h}\)
This does not even make sense. Suggestion: work through the difference quotient algebraically before you start taking limits. See below.
What to do about \(\displaystyle (x + h)^{-1/2}\)
As Halls of Ivy said, you will have to "rationalize" the numerator, ONCE YOU GET IT RIGHT.
Answer should be something like -3x^{-3/2} by using power rule.
Well, no.
First off, you were wise to check your answer by using the power rule. BUT
\(\displaystyle x \ne 0\ and\ f(x) = \dfrac{9}{\sqrt{x}} \implies f(x) = 9x^{-(1/2)}\ and\ f'(x) = 9 * \left(-\dfrac{1}{2}\right) * x^{\{-(1/2) - 1\}} \ne -3x^{-(3/2)}.\)
What is the correct answer using the power rule?
OK Using difference quotients
\(\displaystyle Given:\ f(x) = \dfrac{9}{\sqrt{x}},\ find\ f'(x).\)
\(\displaystyle Step\ 1:\ f(x + h) = \dfrac{9}{\sqrt{x + h}}.\) Easy step.
\(\displaystyle Step\ 2:\ f(x + h) - f(x) = \dfrac{9}{\sqrt{x + h}} - \dfrac{9}{\sqrt{x}}.\) Easy step.
Now you need to do some algebra. What is the difference of two fractions?
\(\displaystyle \dfrac{9}{\sqrt{x + h}} - \dfrac{9}{\sqrt{x}} = \dfrac{9\sqrt{x} - 9\sqrt{x + h}}{\sqrt{x + h} * \sqrt{x}}.\) Easy step.
OK Now Halls told you what the trick is here, namely "rationalize" the numerator. What you are hoping to get is a product where h is a factor.
Once you have your fraction rationalized, it is time for step 3.
\(\displaystyle \dfrac{f(x + h) - f(x)}{h} = what?\)
Use algebra to simplify, and only then do you take the limit
\(\displaystyle \displaystyle \lim_{h \rightarrow 0}\dfrac{f(x + h) - f(x)}{h} = f'(x).\)
