What is the instantaneous rate of change of #f(x)=1/(2x-5)# at #x=1 #?

1 Answer
Truong-Son N.
Jul 15, 2016

The "instantaneous rate of change" is just another way of saying "derivative", or #d/(dx)[f(x)]#, or #(df(x))/(dx)#, or #f'(x)#.

Taking the derivative using the Power Rule asks you to do this:

#stackrel("Power Rule")overbrace(\mathbf(d/(dx)[x^n] = nx^(n-1)))#

So, what we have is:

#d/(dx)[(2x - 5)^(-1)]#

#= stackrel("Power Rule")overbrace((-1)(2x - 5)^(-2)) cdot stackrel("Chain Rule")overbrace(2)#

Don't forget to use the chain rule on composite functions, like #(2x - 5)^(-1)#, which makes you take the derivative of #2x + C# (which is #2 + 0 = 2#).

This is a composite function because we can say #f(x) = x^(-1)# and #g(x) = (2x - 5)#, giving #f(g(x)) = (f @ g)(x) = (2x - 5)^(-1)#.

#=> -2/(2color(red)(x) - 5)^2#

Finally, finding the derivative at #x = 1# is asking, "What is #f'(1)#?". So, really, just plug in #1# to #f'(x)#.

#color(blue)(f'(1)) = -2/(2color(red)((1)) - 5)^2#

#= -2/(-3)^2#

#= color(blue)(-2/9)#

Related questions
See all questions in Instantaneous Rate of Change at a Point
Impact of this question
1762 views around the world
You can reuse this answer
Creative Commons License