We have seen in continuity that if a function is continuous at a point x=a (say). Then its graph is an unbroken curve at x=a and there are no holes and jumps in the graph of the function in the neighbourhood of point x=a
Now a question arises:
What do we mean when we say that a function f(x) is differentiable at a point x=c ?
In the following discussion we shall try to answer the question.
Consider a function f(x) defined on an open interval (a,b)
let P(c,f(c)) be a point on the curve y=f(x) and let the left hand side right hand side respectively of point P as shown in figure
Then the slope of chord PQ
and, slope of chord PR,
We know that tangent to a curve at a point P(say) is the limiting position of chord PQ when . Therefore, as latex \displaystyle h\longrightarrow 0 &s=1$, points Q and R both tends to P from left hand and right hand sides respectively .
Consequently, chord PQ and PR becomes tangents at point P.
Slope of the tangent at point P, which is limiting position of chords drawn on the right hand side of point P.
Now, f(x) is differentiable at x=c
There is a unique tangent at point P.
Thus,f(x) is differentiable at point P, if there exists a unique tangent at point P. In other words f(x) is differentiable at a point P if the curve have P as a corner point.
“i.e. the function is not differentiable at those points on which function has jumps (or holes) and sharp edges.”
Let us consider the function f(x) =|x-1| which can be graphically shown ,
Which shows f(x) is not differentiable at x=1. Since f(x) has sharp edge at x=1.
The right hand derivative at x=1 is 1 and left hand derivative at x=1 is -1 .
Thus f(x) is not differentiable at x=1