Question

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

Answer 1

`2/25`

Explanation:

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

differentiate using the `color(blue)" chain rule "`

`d/dx [ f(g(x)) ] = f'(g(x)) . g'(x)`

Rewrite `1/(x^3 - 2x + 5) = (x^3 - 2x + 5)^-1`
`"------------------------------------------------------"`

f(g(x)) = `(x^3-2x+5)^-1`

`rArr f'(g(x)) = -1(x^3-2x+5)^-2`

and g(x) = `x^3-2x+5 rArr g'(x) = 3x^2-2`
`"----------------------------------------------------------"`
now substitute these values into the derivative

`rArr f'(x) = -(x^3-2x+5)^-2 xx (3x^2-2)`

`= (-(3x^2-2))/(x^3-2x+5)^2`

and f'(0)`=( -(-2))/(5)^2 = 2/25`