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

Jul 15, 2016

The "instantaneous rate of change" is just another way of saying "derivative", or $\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right]$, or $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}$, or $f ' \left(x\right)$.

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

$\stackrel{\text{Power Rule}}{\overbrace{\setminus m a t h b f \left(\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}\right)}}$

So, what we have is:

$\frac{d}{\mathrm{dx}} \left[{\left(2 x - 5\right)}^{- 1}\right]$

$= \stackrel{\text{Power Rule")overbrace((-1)(2x - 5)^(-2)) cdot stackrel("Chain Rule}}{\overbrace{2}}$

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

This is a composite function because we can say $f \left(x\right) = {x}^{- 1}$ and $g \left(x\right) = \left(2 x - 5\right)$, giving $f \left(g \left(x\right)\right) = \left(f \circ g\right) \left(x\right) = {\left(2 x - 5\right)}^{- 1}$.

$\implies - \frac{2}{2 \textcolor{red}{x} - 5} ^ 2$

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

$\textcolor{b l u e}{f ' \left(1\right)} = - \frac{2}{2 \textcolor{red}{\left(1\right)} - 5} ^ 2$

$= - \frac{2}{- 3} ^ 2$

$= \textcolor{b l u e}{- \frac{2}{9}}$