Question

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

Answer 1

`f` is differentiable at `x in RR-{2}, and, f'(x)=-5/(x-2)^2, x in RR-{2}.`

Explanation:

Let `f : RR rarr RR`, be a function and let `x in RR`. We say that the

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

exists `: lim_(trarrx) (f(t)-f(x))/(t-x), where, t in RR, t!=x.`

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

`x`, and, is denoted by `f'(x).`

We have, `f(x)=(2x+1)/(x-2), x in RR-{2}`

`rArr f(t)=(2t+1)/(t-2), t in RR-{2}, tnex`

`:. f(t)-f(x)=(2t+1)/(t-2)-(2x+1)/(x-2)`

`={(2t+1)(x-2)-(2x+1)(t-2)}/((t-2)(x-2))`

`={(2tx-4t+x-2)-(2tx-4x+t-2)}/((t-2)(x-2))`

`=(-5t+5x)/((t-2)(x-2))=(-5(t-x))/((t-2)(x-2))`

Since, `t!=x,` we have, `(f(t)-f(x))/(t-x)=-5/((t-2)(x-2))`

Now, `lim_(trarrx) (f(t)-f(x))/(t-x)=lim_(trarrx) -5/((t-2)(x-2))`

`=-5/((x-2)(x-2))=-5/(x-2)^2`.

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

fun. `f` is differentiable at `x in RR-{2}, and, f'(x)=-5/(x-2)^2, x in RR-{2}.`

Enjoy maths.!