Function $f(x)$ is differentiable at a point P, when there exist a unique tangent at the point P. The common situations in which a function would not be differentiable at a point – vertical tangent lines, discontinuities and sharp turns in the graph.

Consider the function $f(x)$ = $|x|$. The graph of this function is given below

Differentiability Math

The graph shows that $f(x)$ is not differentiable at $x = 0$ as $f(x)$ has sharp edge at $x=0$. The function is continuous at $x =0$ but it is not differentiable there. This clearly says that continuity is not a strong enough condition to guarantee differentiability. On the other hand, when the function is differentiable at a point then it should be continuous at that points. i.e, if a function $f$ is differentiable at $x=c$, hence $x$ is continuous at $x=c$.

The function is known as differentiable function when it's differentiable at domain's each point.
Function $y = f(x)$ is called differentiable in [a, b],
  1. $y = f(x)$ is differentiable at each point of (a, b).
  2. $f'(a+0)$ = $\lim_{h \to 0}$$\frac{f(a+h) – f(a)}{h}$ and $f'(b+0)$ = $\lim_{h \to 0}$ $\frac{f(b-h) – f(b)}{-h}$ both are existing.
  1. Every polynomial function, exponential function and constant function is differentiable at each point of the real line.
  2. If $f(x)$ and $g(x)$ are two functions, the sum, difference, quotient and product of 2 differentiable functions are differentiable.
  3. Logarithmic functions, trigonometric functions and inverse trigonometric functions are differentiable in there domain.
  4. Composition of two differentiable function is also differentiable function.

Solved Example

Question: If a function $f(x)$ is defined as
Differentiabiliy Solved Problems

Check the differentiability of the function at $x = 0$ and $x = 1$.

Solution:
 
At $x = 0$

$Rf'(a)$ = $\lim_{h \to 0}$$\frac{f(a+h) – f(a)}{h}$

$Rf'(0)$= $\lim_{h \to 0}$$\frac{f(h) – f(0)}{h}$

= $\lim_{h \to 0}$$\frac{h^{2}}{h}$ = $0$

$Lf'(0)$ = $\lim_{h \to 0}$$\frac{f(a-h) – f(a)}{-h}$

= $\lim_{h \to 0}$$\frac{f(-h) – f(0)}{-h}$

= $\lim_{h \to 0}$$\frac{-(-h)-0}{-h}$ = $-1$

Since $Rf'(0) \neq Lf'(0)$, \therefore $f'(0)$ dose not exist.

At $x = 1$,

$Rf'(1)$ = $\lim_{h \to 0}$$\frac{f(1+h) – f(1)}{h}$

= $\lim_{h \to 0}$$\frac{f[(1+h)^{2}-(1+h)+1]}{h}$

= $\lim_{h \to 0}$$\frac{h^{2}+h}{h}$

=$1$

$Lf'(1)$ = $\lim_{h \to 0}$$\frac{f(1-h) – f(1)}{(1-h)-1}$

= $\lim_{h \to 0}$$\frac{(1-h)^{2}- 1}{-h}$

= $\lim_{h \to 0}$$\frac{(h)^{2}- 2h}{-h}$

= $lim_{h \to 0}(2-h)$ = $2$

Since $Rf'(1) \neq Lf'(1)$, \therefore $f'(1)$ dose not exist.

\therefore $f(x)$ is not differentiable at $x=0$ and $x=1$.