What is the instantaneous rate of change of f(x)=1/(x^3-2x+5 )f(x)=1x3−2x+5 at x=1 x=1?

1 Answer
Jim G.
Apr 5, 2016

-1/16−116

Explanation:

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

rewrite f(x) as f(x) = (x^3 - 2x + 5)^-1 (x3−2x+5)−1

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

Related questions
See all questions in Instantaneous Rate of Change at a Point
Impact of this question
1481 views around the world
You can reuse this answer
Creative Commons License