Derivative of 1/sqrt(x) via the limit definition

Start by forming the difference quotient: $\frac{f(x+h)-f(x)}{h} = \frac{1}{h}\left(\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}\right) = \frac{1}{h}\cdot\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}.$ This combines the two fractions into one, but it still has an indeterminate $0/0$ form as $h\to 0$, so we need to simplify the numerator further.

Multiply numerator and denominator by $\sqrt{x}+\sqrt{x+h}$ to rationalize: $\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}\cdot\frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} = \frac{x-(x+h)}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})} = \frac{-h}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})}.$ Dividing by $h$ cancels the factor of $h$, leaving $\frac{f(x+h)-f(x)}{h} = \frac{-1}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})}.$

Now let $h \to 0$: the expression is continuous in $h$ at $h=0$ (for $x>0$), so we substitute $h=0$ directly to get $f'(x) = \frac{-1}{\sqrt{x}\cdot\sqrt{x}\cdot(2\sqrt{x})} = \frac{-1}{2x\sqrt{x}}.$ This can also be written as $f'(x) = -\dfrac{1}{2}x^{-3/2}$.

The answer is square roots. Multiplying by the conjugate-style expression eliminated the square root terms from the numerator, turning the difference of square roots into the simple expression negative h. This removed the indeterminate zero over zero form and let the factor of h cancel with the denominator, allowing the limit to be evaluated directly.