How do you find the instantaneous rate of change of the function #f(x)=x/(x+2)# when x=2?

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Answer 1 · 2016-02-07T21:32:04

#f'(2)=1/8#, so the instantaneous rate of change at #x=2# is #1/8#

The instantaneous rate of change at #x=2# is equal to the value of the derivative at #x=2#.

To find the derivative, use the quotient rule, which states that

#d/dx[g(x)/(h(x))]=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2#

Applying this to the function at hand, we see that

#f'(x)=((x+2)d/dx[x]-xd/dx[x+2])/(x+2)^2#

Note that both of these derivatives are equal to #1#.

#f'(x)=((x+2)(1)-x(1))/(x+2)^2#

#f'(x)=(x+2-x)/(x+2)^2#

#f'(x)=2/(x+2)^2#

The instantaneous rate of change at #x=2# is

#f'(2)=2/(2+2)^2=2/4^2=2/16=1/8#