# Differentiability at a Point

Let f (x) be a real valued function defied on an open interval (a, b) and let c ϵ (a, b). Then f (x) is said to be differentiable or derivable at x = c, if $$\underset{x\to c}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( c \right)}{x-c}$$ exist finitely.This limit is called the derivative or differential coefficient of the function f (x) at x = c, and is denoted f’ (c) or Df (c) or {d/dx [f(x)]}ₓ = c. Thus, f (x) is differentiable at x = c

⇒ $$\underset{x\to c}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( c \right)}{x-c}$$.

⇒ $$\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( c \right)}{x-c}=\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( c \right)}{x-c}$$.

⇒ $$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( c-h \right)-f\left( c \right)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( c+h \right)-f\left( c \right)}{h}$$.

The limits $$\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( c \right)}{x-c}$$ or $$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( c-h \right)-f\left( c \right)}{-h}$$ is called the left hand derivative of f (x) at x = c and is denoted by f’ (c) or Lf’ (c), while $$\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( c \right)}{x-c}$$or $$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( c+h \right)-f\left( c \right)}{h}$$ is called the right hand derivative of f (x) at x = c and is denoted by f’ (c⁺) or Rf’(c).

Thus, f (x) is differentiable at x = c ⇒ Lf’ (c) = Rf’(c). If Lf’ (c) ≠ Rf’(c), we say that f (x) is not differentiable at x = c.

Meaning of differentiability of a function at a point: f (x) is differentiable at point P, iff there exists a unique tangent at point P. in order words, f (x) is differentiable at a point P iff the curve does not have P as a corner point.

Consider the function f (x) = |x|. This function is not differentiable at x = 0, because if we draw tangent at the origin as the limiting position of the chords on the left hand side of the origin, it is the line y = – x whereas the tangent at the origin as the limiting position of the chords on the right hand side of the origin is the line y = x. mathematically, left hand derivative at the origin is – 1 (slope of the line y = – x) and the fight hand derivate at the origin is 1 (slope of the line y = x).

Let f (x) be a differentiable function at a point P. then the curve y = f (x) has a unique tangent at P. since tangent at P is the limiting position of the chord PQ when Q → P. So, if f (x) is differentiable at a point P, then chords exist on both sides of P. consequently f (x) is continuous at P.

It follows from the above discussion that, if a function is not differentiable at x = c, then either it has (c, f (c)) as a corner point or it is discontinuous at x = c.

Also, every differentiable function is continuous.

Relation between continuity and differentiability: In the above discussion, we have observed that if a function is differentiable at a point, then it should be continuous at that point and a discontinuous function cannot be differentiable. This fact is provided in the following theorem.

If a function I differentiable at a point, it is necessarily continuous at that point. But the converse is not necessarily true.

Or

f (x) is differentiable at x = c ⇒ f (x) is continuous at x = c.

Converse:  The converse of the above theorem is not necessarily true i.e. a function may be continuous at a point but may not be differentiable at that point because f (x) = |x| is continuous at x = 0 but it is not differentiable at x = 0.

Evaluate:  show that f (x) = |x| is not differentiable at x = 0.

Solution: we have,

(LHD at x = 0)

= $$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( 0 \right)}{x-0}$$.

= $$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 0-h \right)-f\left( 0 \right)}{0-h-0}$$.

= $$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 0-h \right)-f\left( 0 \right)}{-h}$$.

= $$\underset{h\to 0}{\mathop{\lim }}\,\frac{\left| h \right|-\left| 0 \right|}{-h}$$.

= $$\underset{h\to 0}{\mathop{\lim }}\,\frac{\left| -h \right|}{-h}$$.

$$\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{-h}=\underset{h\to 0}{\mathop{\lim }}\,-1=-1$$.

= $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( 0 \right)}{x-0}$$.

= $$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 0+h \right)-f\left( 0 \right)}{h}$$.

= $$\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( h \right)-f\left( 0 \right)}{h}$$.

= $$\underset{h\to 0}{\mathop{\lim }}\,\frac{\left| h \right|-\left| 0 \right|}{h}$$.

$$\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{h}=\underset{h\to 0}{\mathop{\lim }}\,1=1$$.

∴ (LHD at x = 0) ≠ (RHD at x = 0).

So, f (x) is not differentiable at x = 0.