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# Introduction of Differentiability of a Function at a point

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 $\displaystyle \Rightarrow \frac { f\left( c-h \right) -f\left( c \right) }{ \left( c-h \right) -\left( c \right) } \\ \Rightarrow \frac { f\left( c-h \right) -f\left( c \right) }{ -h }$

and, slope of chord PR, $\displaystyle \Rightarrow \frac { f\left( c-h \right) -f\left( c \right) }{ \left( c+h \right) -\left( c \right) } \\ \Rightarrow \frac { f\left( c-h \right) -f\left( c \right) }{ h }$

We know that tangent to a curve at a point P(say) is the limiting position of chord PQ when $\displaystyle Q\longrightarrow P$. 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.

Thus , $\displaystyle \lim _{ h\longrightarrow 0 }{ \frac { f\left( c-h \right) f\left( c \right) }{ -h } } =\lim _{ h\longrightarrow 0 }{ \left( slope\quad of\quad chord\quad PQ \right) } \\ =\lim _{ Q\longrightarrow P }{ \left( slope\quad of\quad chord\quad PQ \right) }$

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 $\displaystyle \Rightarrow \lim _{ h\longrightarrow 0 }{ \frac { f\left( c-h \right) f\left( c \right) }{ -h } } =\lim _{ h\longrightarrow 0 }{ \frac { f\left( c+h \right) f\left( c \right) }{ h } }$

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.

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

April 18, 2019
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