# How do you prove from the definition of differentiability that the function f(x)=(2x+1)/(x-2) is differentiable?

Sep 28, 2016

$f$ is differentiable at $x \in \mathbb{R} - \left\{2\right\} , \mathmr{and} , f ' \left(x\right) = - \frac{5}{x - 2} ^ 2 , x \in \mathbb{R} - \left\{2\right\} .$

#### Explanation:

Let $f : \mathbb{R} \rightarrow \mathbb{R}$, be a function and let $x \in \mathbb{R}$. We say that the

function $f$ is Differentiable at $x$ , if, the following Limit

exists $: {\lim}_{t \rightarrow x} \frac{f \left(t\right) - f \left(x\right)}{t - x} , w h e r e , t \in \mathbb{R} , t \ne x .$

Also, if the above Limit exists, it is called the Derivative of $f$ at

$x$, and, is denoted by $f ' \left(x\right) .$

We have, $f \left(x\right) = \frac{2 x + 1}{x - 2} , x \in \mathbb{R} - \left\{2\right\}$

$\Rightarrow f \left(t\right) = \frac{2 t + 1}{t - 2} , t \in \mathbb{R} - \left\{2\right\} , t \ne x$

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

$= \frac{\left(2 t + 1\right) \left(x - 2\right) - \left(2 x + 1\right) \left(t - 2\right)}{\left(t - 2\right) \left(x - 2\right)}$

$= \frac{\left(2 t x - 4 t + x - 2\right) - \left(2 t x - 4 x + t - 2\right)}{\left(t - 2\right) \left(x - 2\right)}$

$= \frac{- 5 t + 5 x}{\left(t - 2\right) \left(x - 2\right)} = \frac{- 5 \left(t - x\right)}{\left(t - 2\right) \left(x - 2\right)}$

Since, $t \ne x ,$ we have, $\frac{f \left(t\right) - f \left(x\right)}{t - x} = - \frac{5}{\left(t - 2\right) \left(x - 2\right)}$

Now, ${\lim}_{t \rightarrow x} \frac{f \left(t\right) - f \left(x\right)}{t - x} = {\lim}_{t \rightarrow x} - \frac{5}{\left(t - 2\right) \left(x - 2\right)}$

$= - \frac{5}{\left(x - 2\right) \left(x - 2\right)} = - \frac{5}{x - 2} ^ 2$.

Thus, we find that, the Limit in question exists, and so, the given

fun. $f$ is differentiable at $x \in \mathbb{R} - \left\{2\right\} , \mathmr{and} , f ' \left(x\right) = - \frac{5}{x - 2} ^ 2 , x \in \mathbb{R} - \left\{2\right\} .$

Enjoy maths.!