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

##### 1 Answer

The "instantaneous rate of change" is just another way of saying "derivative", or

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

This is a composite function because we can say

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

Finally, finding the derivative at

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

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

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