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

$$\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h}.$$

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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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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