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.

  • 2
    $\begingroup$ 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. $\endgroup$ – André Nicolas Dec 12 '13 at 3:24
  • $\begingroup$ Alternately, from the first expression I gave, use $\frac{a-b}{a^2-b^2}=\frac{1}{a+b}$ to simplify $\frac{\sqrt{x}-\sqrt{c}}{x-c}$. $\endgroup$ – André Nicolas Dec 12 '13 at 3:31
  • 1
    $\begingroup$ @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. $\endgroup$ – Stefan Smith Dec 12 '13 at 3:34


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

Multiplying by the numerator's conjugate

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

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

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


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