# How do you find the derivative 1/(x^7 + 2)?

May 24, 2015

I would usr the Quotient Rule as:
$f ' \left(x\right) = \frac{- 7 {x}^{6}}{{x}^{7} + 2} ^ 2$

Remember the Quotient Rule to derive a function $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$
you get:
$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

May 24, 2015

To find the derivative of $f \left(x\right) = \frac{1}{{x}^{7} + 2}$ you would use the quotient rule:

$f \left(x\right) = g \frac{x}{h \left(x\right)}$ then $f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h {\left(x\right)}^{2}}$

Thus:

let $g \left(x\right) = 1$ so $g ' \left(x\right) = 0$

let $h \left(x\right) = \left({x}^{7} + 2\right)$ so $h ' \left(x\right) = 7 {x}^{6}$

Therefore we get that:

$f ' \left(x\right) = \frac{- 7 {x}^{6}}{{x}^{7} + 2} ^ 2$

this shows us that $f \left(x\right)$ is always decreasing, as its derivative is always negative

May 25, 2015

$f \left(x\right) = \frac{1}{{x}^{7} + 2} = {\left({x}^{7} + 2\right)}^{-} 1$
$f ' \left(x\right) = - 1 {\left({x}^{7} + 2\right)}^{-} 2 \cdot 7 {x}^{6}$
$= \frac{- 7 {x}^{6}}{{x}^{7} + 2} ^ 2$