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.!
