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

1 Answer
Jim G.
Apr 15, 2016

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

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