# How do you find the derivative of f(x)=4/(sqrtx-5) using the limit definition?

Dec 21, 2017

$f ' \left(x\right) = \frac{- 2}{\left(\sqrt{x} - 5\right) \left(\sqrt{x} - 5\right) \left(\sqrt{x}\right)}$

#### Explanation:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\frac{4}{\sqrt{x + h} - 5} - \frac{4}{\sqrt{x} - 5}}{h}$ (difference quotient)

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\frac{4 \left(\sqrt{x} - 5\right)}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right)} - \frac{4 \left(\sqrt{x + h} - 5\right)}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right)}}{h}$ (combine numerator into 1 fraction)

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{4 \sqrt{x} - 20 - 4 \sqrt{x + h} + 20}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right) h}$ (multiply denominator of the fraction in the numerator to the denominator of the larger fraction)

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{4 \sqrt{x} - 4 \sqrt{x + h}}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right) h}$

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\left(4 \sqrt{x} - 4 \sqrt{x + h}\right) \left(4 \sqrt{x} + 4 \sqrt{x + h}\right)}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right) h \left(4 \sqrt{x} + 4 \sqrt{x + h}\right)}$ (conjugate)

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{16 x - 16 x - 16 h}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right) h \left(4 \sqrt{x} + 4 \sqrt{x + h}\right)}$

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{- 16 h}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right) h \left(4 \sqrt{x} + 4 \sqrt{x + h}\right)}$

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{- 16}{\left(\sqrt{x + h} - 5\right) \left(\sqrt{x} - 5\right) \left(4 \sqrt{x} + 4 \sqrt{x + h}\right)}$ (remove h from both numerator and denominator)

now you can use direct substitution:
$f ' \left(x\right) = \frac{- 16}{\left(\sqrt{x + 0} - 5\right) \left(\sqrt{x} - 5\right) \left(4 \sqrt{x} + 4 \sqrt{x + 0}\right)}$

$f ' \left(x\right) = \frac{- 16}{\left(\sqrt{x} - 5\right) \left(\sqrt{x} - 5\right) \left(4 \sqrt{x} + 4 \sqrt{x}\right)}$

$f ' \left(x\right) = \frac{- 16}{\left(\sqrt{x} - 5\right) \left(\sqrt{x} - 5\right) \left(8 \sqrt{x}\right)}$

$f ' \left(x\right) = \frac{- 2}{\left(\sqrt{x} - 5\right) \left(\sqrt{x} - 5\right) \left(\sqrt{x}\right)}$