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If $f(x) = 1/\sqrt{x}$, find $f^{\prime}(x)$. Please show the answer by using


I know the answer by using the shortcut, but my teacher wants me to get the answer using that equation. I am stuck and I got to: $-\sqrt{x}/2x^2$ by using the $\frac{f(x+h)-f(x)}{h}$.

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Please use LaTeX formatting for mathematical expressions - makes your question easier to read :) – Johnny Westerling Sep 25 '12 at 18:18
Sorry, I'm new to this site. – dsta Sep 25 '12 at 18:19

3 Answers 3

Step 1. Plug in $f(x+h) = \frac{1}{\sqrt{x + h}}$ to the above.

Step 2. The limit definition of the derivative will have numerator involving two fractions. Get a common denominator and add them.

Step 3. The numerator, at this point, will be a fraction that has $\sqrt{x} - \sqrt{x + h}$ in its numerator. Multiply the numerator and denominator by the conjuage $\sqrt{x} + \sqrt{x+h}$ and simplify everything. You will be able to cancel the $h$ in the denominator at this point.

Hopefully you can finish it at this point.

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Write the equation and take the LCM in Numerator($f(x+h)-f(x)$) and now rationalize it. Now cancel h and then apply limit.

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$$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\lim_{ h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}=\lim_{ h \to 0}\frac{\sqrt x -\sqrt{x+h}}{\sqrt{x(x+h)}h}=$$

$$=\lim_{ h \to 0}\frac{x -(x+h)}{\sqrt{x(x+h)}h(\sqrt x +\sqrt{x+h})}=-\lim_{ h \to 0}\;\;\frac{h}{h} \frac{1}{\sqrt{x(x+h)}(\sqrt x +\sqrt{x+h})}=$$

$$=-\frac 1 2 \frac{1}{x^{\frac 3 2}}\,\,,\,(\text{ as}\,\,\, h \ne 0\,\,\,\text{ as}\,\,\lim_{ h \to 0})=-\frac{\sqrt x}{2x^2}$$

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Notice the homework tag. – Graphth Sep 25 '12 at 18:23
@Graphth, could you please tell me its implication ? – lab bhattacharjee Sep 25 '12 at 18:24
We shouldn't do the homework for them. That's the whole point of the tag. If we are to treat these problems exactly as if they are not homework, then there's no reason to have a homework tag. – Graphth Sep 25 '12 at 18:27
After thinking about this some, I do want to point out that I have in the past given full solutions to homework questions. But, I am trying to just do hints now because that's probably what is best. – Graphth Sep 27 '12 at 12:50
@Graphth, I've also started doing that… – lab bhattacharjee Sep 27 '12 at 12:52

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