# How do I use the limit definition of derivative to find f'(x) for f(x)=1/sqrt(x) ?

Oct 1, 2014

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

Note that a square root is equivalent to raising an expression to the $\frac{1}{2}$ power.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$f \left(x + h\right) = \frac{1}{\sqrt{x + h}}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h}$

Find the common denominator

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} \cdot \frac{\sqrt{x}}{\sqrt{x}} - \frac{1}{\sqrt{x}} \cdot \frac{\sqrt{x + h}}{\sqrt{x + h}}}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{\sqrt{x}}{\sqrt{x} \sqrt{x + h}} - \frac{\sqrt{x + h}}{\sqrt{x} \sqrt{x + h}}}{h}$

Consolidate the numerator of the complex fraction.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x} \sqrt{x + h}}}{h}$

Dividing fractions is equivalent to multiplying by the reciprocal

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{1}{h} \cdot \left(\frac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x} \sqrt{x + h}}\right)$

Rationalize the numerator

f'(x)=lim_(h->0) 1/h*((sqrt(x)-sqrt(x+h))/(sqrt(x)sqrt(x+h)))*(sqrt(x)+sqrt(x+h))/(sqrt(x)+sqrt(x+h)

Simplify. Remember difference of perfect squares.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{1}{h} \cdot \left(\frac{x - \left(x + h\right)}{\sqrt{x} \sqrt{x + h} \left(\sqrt{x} + \sqrt{x + h}\right)}\right)$

Distribute the negative in the numerator

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{1}{h} \cdot \frac{x - x - h}{\sqrt{x} \sqrt{x + h} \left(\sqrt{x} + \sqrt{x + h}\right)}$

The $x ' s$ resolve to zero.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{1}{h} \cdot \frac{- h}{\sqrt{x} \sqrt{x + h} \left(\sqrt{x} + \sqrt{x + h}\right)}$

The $h ' s$ can be cancelled.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{- 1}{\sqrt{x} \sqrt{x + h} \left(\sqrt{x} + \sqrt{x + h}\right)}$

Now we can substitute in 0 for $h$.

$\frac{- 1}{\sqrt{x} \sqrt{x + 0} \left(\sqrt{x} + \sqrt{x + 0}\right)}$

$\frac{- 1}{\sqrt{x} \sqrt{x} \left(\sqrt{x} + \sqrt{x}\right)}$

Manipulate the exponents

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