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} $

Also Read

→Limits of a Function
→Continuity of a Function
→Differentiability of a Function
→Rules for differentiation
→Higher order derivatives

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