# What is the instantaneous rate of change of f(x)=1/(x^3-2x+5 ) at x=1 ?

Apr 5, 2016

$- \frac{1}{16}$

#### Explanation:

This is the value of the derivative at x = 1.

rewrite f(x) as f(x) = ${\left({x}^{3} - 2 x + 5\right)}^{-} 1$

now differentiate using the $\textcolor{b l u e}{\text{ chain rule }}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$
$\text{-----------------------------------------------------}$

f(g(x)) = ${\left({x}^{3} - 2 x + 5\right)}^{-} 1$

$\Rightarrow f ' \left(g \left(x\right)\right) = - {\left({x}^{3} - 2 x + 5\right)}^{-} 2$

and g(x) $= {x}^{3} - 2 x + 5 \Rightarrow g ' \left(x\right) = 3 {x}^{2} - 2$
$\text{--------------------------------------------------------}$

$\Rightarrow f ' \left(x\right) = - {\left({x}^{3} - 2 x + 5\right)}^{-} 2 . \left(3 {x}^{2} - 2\right)$

$= \frac{- \left(3 {x}^{2} - 2\right)}{{x}^{3} - 2 x + 5} ^ 2$

$\Rightarrow f ' \left(1\right) = \frac{- \left(3 - 2\right)}{1 - 2 + 5} ^ 2 = \frac{- 1}{4} ^ 2 = - \frac{1}{16}$