# What is the instantaneous rate of change of f(x)=1/(x^3-2x+5 ) at x=0 ?

Apr 15, 2016

$\frac{2}{25}$

#### Explanation:

This is just the value of the derivative at x = 0.

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

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

Rewrite $\frac{1}{{x}^{3} - 2 x + 5} = {\left({x}^{3} - 2 x + 5\right)}^{-} 1$
$\text{------------------------------------------------------}$

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

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

and g(x) = ${x}^{3} - 2 x + 5 \Rightarrow g ' \left(x\right) = 3 {x}^{2} - 2$
$\text{----------------------------------------------------------}$
now substitute these values into the derivative

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

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

and f'(0)$= \frac{- \left(- 2\right)}{5} ^ 2 = \frac{2}{25}$