Question

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

Answer 1

`-1/16`

Explanation:

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

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

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

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

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

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

and g(x) `= x^3-2x+5 rArr g'(x) = 3x^2 - 2`
`"--------------------------------------------------------"`

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

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

`rArr f'(1) = (-(3-2))/(1-2+5)^2 = (-1)/(4)^2 = -1/16`