Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to use the basic definition of the derivative to find the derivative of $$f(x)=\frac{1}{\sqrt{x}}$$ for x>0

I need to use the limit... $$\lim_{x \to c}\frac{f(x)-f(c)}{x-c}$$ So I have $$\lim_{x \to c}\frac{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{c}}}{x-c}$$

I just keep getting stuck and maybe I am just simplifying incorrectly-

Any advice would be great! Thanks.

share|improve this question
Good so far. Then bring to a common denominator and get $\frac{\sqrt{c}-\sqrt{x}}{\sqrt{x}\sqrt{c}(x-c)}$. Now multiply top and bottom by $\sqrt{c}+\sqrt{x}$. Alternately, from your expression as it stands, multiply top and bottom by $\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{c}}$ and do some algebra. –  André Nicolas Dec 12 '13 at 3:24
Alternately, from the first expression I gave, use $\frac{a-b}{a^2-b^2}=\frac{1}{a+b}$ to simplify $\frac{\sqrt{x}-\sqrt{c}}{x-c}$. –  André Nicolas Dec 12 '13 at 3:31
@user2553807 : in my opinion, it's easier to use the equivalent form $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$. The reason is that it tends to be easier to simplify algebraically. Using your version, you will have to factor out a factor of $x-c$ from a numerator, which in some problems (not necessarily this problem) may be difficult. –  Stefan Smith Dec 12 '13 at 3:34

2 Answers 2

Notice that:


share|improve this answer


$$f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt x}}{h} = \frac{\frac{\sqrt x - \sqrt{x+h}}{\sqrt{x+h}\sqrt x}}{h}$$

Multiplying by the numerator's conjugate

$$\frac{\sqrt x - \sqrt{x+h}}{\sqrt{x+h}\sqrt x} \cdot \frac{\sqrt x + \sqrt{x+h}}{\sqrt x +\sqrt{x+h}} = \frac{- h}{\left(\sqrt x +\sqrt{x+h}\right)\sqrt{x+h}\sqrt x}$$

$$ \lim_{h\to0} \frac{- h}{h\left(\sqrt x +\sqrt{x+h}\right)\sqrt{x+h}\sqrt x}$$ $$= \lim_{h\to0} \frac{- 1}{\left(\sqrt x +\sqrt{x+h}\right)\sqrt{x+h}\sqrt x}$$

$$= \frac{- 1}{\left(\sqrt x +\sqrt{x}\right)\sqrt{x}\sqrt x} = \frac{-1}{2\sqrt x \cdot x} = \frac{-1}{2x^{\frac{3}{2}}}$$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.