# Differentiability of Function

In this section we deal with yet another concept, namely the differentiability of function.

Before we move on to the precise definition of differentiability, we first define a few useful terms that will be subsequently used in the remaining part of this section.

Increment: Let y = f (x) be a given function. It is normal to accept that with a change in the value of independent variable , x in this case , there will also be a change in the corresponding values of the dependent variable y in this case.

Say xi and xf be the initial and final values of independent variable.

The quantity xf − xi is called the increment in the variable x and is usually denoted by Δx.

If Δx be the increment in the value of independent variable, the corresponding change in the value of dependent variable is obtained as follows ,

Δy = yf − yi = f (x + Δx) − f (x)

Clearly the increment may be positive negative or may even be zero.

### Differential Coefficient:

Differential coefficient of y = f (x), with respect to x is defined as the limiting value of the ratio as Δx tends to zero.

It is usually denoted by $\large \frac{dy}{dx} = f'(x)$ , Symbolically

$\large \frac{dy}{dx} = \lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x}$

$\large =\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

$\large =\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$

This formulae is also called the first principle of calculating the differential coefficient of any given function.

### Differentiability of a function at a given point:

The function y = f (x) is said to be differentiable at a general point x , if and only if

$\large \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$ exists finitely , else it is said to be non-differentiable at that point.

It simply means that if y = f (x) has finite differential coefficient at a general point x i.e. f ‘ (x) exists finitely, f (x) is said to be differentiable, else non – differentiable.

Clearly , $\large \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$   will exists finitely if

$\large \lim_{h \rightarrow 0^-}\frac{f(x-h)-f(x)}{-h}$ and $\large \lim_{h \rightarrow 0^+}\frac{f(x+h)-f(x)}{h}$   are equal and finite.

These limiting values are called left hand derivative and right hand derivative respectively of the function y = f (x) and usually denoted by f ‘ (x − 0) and f ‘ (x + 0) respectively.

Thus, we conclude that

f ‘(x) exists ⇔ f ‘ (x − 0) = f ‘ (x + 0) = some finite quantity.

Before we move on to the geometrical aspect of differentiability, we will provide some illustrations to make the above discussed definition more clear.

### Differentiation based on first Principle

Example : Using the first principle find the differential coefficient of y = f (x), where f (x) is

(i) ex

(ii) tan x

(iii)$\frac{sinx}{x}$

(iv) tan−1 x

Solution :

(i) y = f (x) = ex

we have $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{e^{x+h}-e^x}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{e^x(e^h -1)}{h}$

$\displaystyle = e^x \lim_{h\rightarrow 0} \frac{(e^h -1)}{h}$

$\displaystyle = e^x \times 1$

= ex

(ii) y = f (x) = tan x

We have , $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) – f(x)}{h}$

$\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{tan(x+h) – tanx }{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{\frac{sin(x+h)}{cos(x+h)} – \frac{sinx}{cosx}}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{sin(x+h)cosx-cos(x+h)sinx}{h cos(x+h)cosx}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{sin(x+h-x)}{h cos(x+h)cosx}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{sinh}{h cos(x+h)cosx}$

$\displaystyle =\frac{1}{cos^2 x}$

= sec2x

(iii) y = f (x) = sinx/x

We have , $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{\frac{sin(x+h)}{x+h} – \frac{sinx}{x}}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x sin(x+h) – (x+h)sinx}{h (x+h)x}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x (sin(x+h) -sinx )-h sinx}{h (x+h)x}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x (2 cos(x+h/2) .sinh/2 )-h sinx}{h (x+h)x}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{x (cos(x+h/2 ) .sinh/2 )}{(h/2) (x+h)x}-\lim_{h\rightarrow 0} \frac{sinx}{(x+h)x}$

$\displaystyle = \frac{x cosx}{x^2}-\frac{sinx}{x^2}$

$\displaystyle = \frac{x cosx -sinx}{x^2}$

(iv) y = f (x) = tan−1 x

we have , $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{tan^{-1}(x+h)- tan^{-1}x}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{tan^{-1}\frac{x+h-x}{1+x(x+h)}}{h}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{tan^{-1}\frac{h}{1+x(x+h)}}{\frac{h(1+x(x+h))}{1+x(x+h)}}$

$\displaystyle = \lim_{h\rightarrow 0} \frac{tan^{-1}\frac{h}{1+x(x+h)}}{\frac{h)}{1+x(x+h)}} \times \lim_{h\rightarrow 0} \frac{1}{(1+x(x+h)}$

$\displaystyle = \frac{1}{1+x^2}$