**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 PQÂ Â

and, slope of chord PR,Â

We know that tangent to a curve at a point P(say) is the limiting position of chord PQ whenÂ . Therefore, asÂ 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.

Thus ,Â Â

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 functionÂ 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.

**Mathematically:**

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