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)

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