# How do you find the derivative using the difference quotient f(x) = 5 / (x^2 + 7)?

Apr 12, 2015

The quotient rule is:

$y = f \frac{x}{g} \left(x\right) \Rightarrow y ' = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$.

So:

$y ' = \frac{0 \cdot \left({x}^{2} + 7\right) - 5 \cdot 2 x}{{x}^{2} + 7} ^ 2 = \frac{- 10 x}{{x}^{2} + 7} ^ 2$.

Apr 12, 2015

The details depend on which difference quotient you use :

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

or,

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

For $f \left(x\right) = \frac{5}{{x}^{2} + 7}$ and using the second form:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\frac{5}{{\left(x + h\right)}^{2} + 7} - \frac{5}{{x}^{2} + 7}}{h}$.

$= {\lim}_{h \rightarrow 0} \frac{5 \left({x}^{2} + 7\right) - 5 \left({\left(x + h\right)}^{2} + 7\right)}{h \left({\left(x + h\right)}^{2} + 7\right) \left({x}^{2} + 7\right)}$

$= {\lim}_{h \rightarrow 0} \frac{5 {x}^{2} + 35 - 5 {x}^{2} - 10 x h - 5 {h}^{2} - 35}{h \left({\left(x + h\right)}^{2} + 7\right) \left({x}^{2} + 7\right)}$

$= {\lim}_{h \rightarrow 0} \frac{- 10 x - 5 h}{\left({\left(x + h\right)}^{2} + 7\right) \left({x}^{2} + 7\right)}$

$= \frac{- 10 x}{{\left({x}^{2} + 7\right)}^{2}}$