# How do you find the instantaneous rate of change for f(x)=(x^2-2)/(x-1) for x=2?

Feb 2, 2018

Please refer to the explanation below.

#### Explanation:

We have to take the derivative of the function $f \left(x\right)$, and then plug in $x = 2$ into $f ' \left(x\right)$.

$f \left(x\right) = \frac{{x}^{2} - 2}{x - 1}$

So, we have to use the quotient rule, which states that,

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$

Here, $u = {x}^{2} - 2$, $v = x - 1$

$\therefore f ' \left(x\right) = \frac{2 x \left(x - 1\right) - 1 \cdot \left({x}^{2} - 2\right)}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{2 {x}^{2} - 2 x - {x}^{2} + 2}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} - 2 x + 2}{x - 1} ^ 2$

Plugging in $x = 2$, we get

$f ' \left(2\right) = \frac{{2}^{2} - 2 \cdot 2 + 2}{2 - 1} ^ 2$

$f ' \left(2\right) = \frac{4 - 4 + 2}{1} ^ 2$

$f ' \left(2\right) = \frac{2}{1}$

$f ' \left(2\right) = 2$

So, the instantaneous rate of change will be $2$.

Feb 2, 2018

$m = 2$

#### Explanation:

By finding the derivativ of $f \left(x\right)$

$f \left(x\right) = \frac{{x}^{2} - 2}{x - 1}$
$f ' \left(x\right) = \frac{\left({x}^{2} - 2\right) \cdot 1 - \left(x - 1\right) \cdot 2 x}{x - 1}$

Now we determinate the value of the divertive at the point x=2

$f ' \left(2\right) = 2$