# Question #8a726

Jan 8, 2018

See below.

#### Explanation:

To find the instantaneous rate of change, we need to find the derivative of $\frac{1}{x}$. Putting in values of $x$ will give us the gradient of the tangent at that point. This is different from the average rate of change, which is the gradient of the secant line that joins $2$ points.

So the instantaneous rate of change is given by $f ' \left(a\right)$, and the average rate of change on an interval $\left[a , b\right]$ is given by

$\frac{\left(f \left(b\right)\right) - \left(f \left(a\right)\right)}{b - a}$

$\frac{1}{x} = {x}^{-} 1$

Using the power rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{-} 1\right) = - {x}^{- 2} = - \frac{1}{x} ^ 2$

$\therefore$

${x}_{0} = 2$

$f ' \left({x}_{0}\right) = - \frac{1}{2} ^ 2 = \textcolor{b l u e}{- \frac{1}{4}}$

${x}_{1} = 3$

$f ' \left({x}_{1}\right) = - \frac{1}{3} ^ 2 = \textcolor{b l u e}{- \frac{1}{9}}$

Instantaneous Rate of Change:

Average Rate of Change: