# How do you use the definition of a derivative to find the derivative of f(x) = 4 / sqrt( 5 - x )?

Mar 31, 2018

d/dx( 4/sqrt(5-x)) =2/((5-x)^(3/2)

#### Explanation:

Using the definition of derivative:

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

we have:

$\frac{d}{\mathrm{dx}} \left(\frac{4}{\sqrt{5 - x}}\right) = {\lim}_{h \to 0} \frac{1}{h} \left(\frac{4}{\sqrt{5 - x - h}} - \frac{4}{\sqrt{5 - x}}\right)$

$\frac{d}{\mathrm{dx}} \left(\frac{4}{\sqrt{5 - x}}\right) = {\lim}_{h \to 0} \frac{4}{h} \frac{\sqrt{5 - x} - \sqrt{5 - x - h}}{\sqrt{5 - x - h} \sqrt{5 - x}}$

Rationalize the numerator by multiplying and dividing by:
$\left(\sqrt{5 - x} + \sqrt{5 - x - h}\right)$ and using the algebraic identity:
$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

$\frac{d}{\mathrm{dx}} \left(\frac{4}{\sqrt{5 - x}}\right) = {\lim}_{h \to 0} \frac{4}{h} \frac{\sqrt{5 - x} - \sqrt{5 - x - h}}{\sqrt{5 - x - h} \sqrt{5 - x}} \times \frac{\sqrt{5 - x} + \sqrt{5 - x - h}}{\sqrt{5 - x} + \sqrt{5 - x - h}}$

d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/h((5-x) -(5-x-h)) /(sqrt(5-x-h) sqrt(5-x)(sqrt(5-x) + sqrt(5-x-h))

d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/h(cancel(5)-cancel(x) -cancel(5)+cancel(x)+h) /(sqrt(5-x-h) sqrt(5-x)(sqrt(5-x) + sqrt(5-x-h))

d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/cancel(h)cancel(h) /(sqrt(5-x-h) sqrt(5-x)(sqrt(5-x) + sqrt(5-x-h))

d/dx( 4/sqrt(5-x)) = 4/(sqrt(5-x) sqrt(5-x)(sqrt(5-x) + sqrt(5-x))

$\frac{d}{\mathrm{dx}} \left(\frac{4}{\sqrt{5 - x}}\right) = \frac{4}{2 \left(5 - x\right) \sqrt{5 - x}}$

d/dx( 4/sqrt(5-x)) =2/((5-x)^(3/2)