# How do you differentiate f(x)=1/sqrt(x-4) using first principles?

Nov 22, 2016

$f \left(x\right) = \frac{1}{\sqrt{x - 4}}$

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

$= {\lim}_{h \rightarrow 0} \frac{\frac{1}{\sqrt{\left(x - 4\right) + h}} - \frac{1}{\sqrt{\left(x - 4\right)}}}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\frac{\sqrt{x - 4} - \sqrt{\left(x - 4\right) + h}}{\sqrt{\left(x - 4\right) + h} \sqrt{\left(x - 4\right)}}}{\frac{h}{1}}$

$= {\lim}_{h \rightarrow 0} \frac{\sqrt{x - 4} - \sqrt{\left(x - 4\right) + h}}{\sqrt{\left(x - 4\right) + h} \sqrt{\left(x - 4\right)}} \cdot \frac{1}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\left(\sqrt{x - 4} - \sqrt{\left(x - 4\right) + h}\right)}{h \sqrt{\left(x - 4\right) + h} \sqrt{\left(x - 4\right)}} \cdot \frac{\left(\sqrt{x - 4} + \sqrt{\left(x - 4\right) + h}\right)}{\left(\sqrt{x - 4} + \sqrt{\left(x - 4\right) + h}\right)}$

$= {\lim}_{h \rightarrow 0} \frac{\left(x - 4\right) - \left(\left(x - 4\right) + h\right)}{h \sqrt{\left(x - 4\right) + h} \sqrt{\left(x - 4\right)} \left(\sqrt{x - 4} + \sqrt{\left(x - 4\right) + h}\right)}$

$= {\lim}_{h \rightarrow 0} \frac{- {\cancel{h}}^{1}}{\cancel{h} \sqrt{\left(x - 4\right) + h} \sqrt{\left(x - 4\right)} \left(\sqrt{x - 4} + \sqrt{\left(x - 4\right) + h}\right)}$

$= {\lim}_{h \rightarrow 0} \frac{- 1}{\sqrt{\left(x - 4\right) + h} \sqrt{\left(x - 4\right)} \left(\sqrt{x - 4} + \sqrt{\left(x - 4\right) + h}\right)}$

$= \frac{- 1}{\sqrt{\left(x - 4\right) + 0} \sqrt{\left(x - 4\right)} \left(\sqrt{x - 4} + \sqrt{\left(x - 4\right) + 0}\right)}$

$= \frac{- 1}{\left(x - 4\right) \left(2 \sqrt{x - 4}\right)} = \frac{- 1}{2 \left(x - 4\right) \sqrt{x - 4}} = \frac{- 1}{2 {\left(x - 4\right)}^{\frac{3}{2}}}$