# Differentiability

## Differentiability

Before introducing the concept and condition of differentiability, it is important to know differentiation and the concept of differentiation.

Differential coefficient of a function y= f(x) is written as

d/dx[f(x)] or f' (x) or f (1) (x) and is defined by

f'(x)= limh→0 (f(x+h)-f(x))/h

f'(x) represents nothing but ratio by which f(x) changes for small change in x and can be understood as

f'(x) = lim?x→0 (? y)/(? x) = dy/dx

Then f'(x) represents the rate of change of y w.r.t. x or in other words f' (x) represents slope of the tangent drawn at point 'x' of the curve f(x).

Let us understand the geometrical meaning of differentiation:

Slope of PQ = yQ - yp /yQ

= ((y+?y)-y)/((x+?x)-x) or (f(x+?x)-f(x))/((x+?x)-x)

= ?y/?x = (f(x+?x)-f(x))/?x

Let point Q approach point P, which the curve = y= f(x) i.e. ? x → 0.

Then, we observe graphically that the slope of chord PQ becomes the slope of the tangent at the point P. which is written as dx/dy or f'(x).

Since, point Q is approaching point P from the Right Hand side, we obtain f'(x+) as follows:

f'(x+)= lim?x→0  (f(x+?x)-f(x))/?x

(Right hand derivative)

Similarly,

f' (x-) = lim?x→0 (f(x + ? x)-f(x))/(-? x),?x>0

(Left hand derivative)

Note: For a function to be differentiable at x=a, we should have f'(a-)=f'(a+)

i.e. limh→0  (f(a-h)-f(a))/(-h) = limh→0  (f(a+h)-f(a))/h

From the above graphs, one must not infer that a curve is non-differentiable only at points of discontinuity. Non differentiability conditions also arise when the curve is continuous and the curve suddenly changes direction. The easiest example of a curve being continuous and non-differentiable is y=|x| at x=0. However when there is a smooth change or gradual change in slope or trajectory of curve, the derivative exists.

Few more illustrations are given below:

Refer to the following graphs:

In figure (i), f'(a) exists and is finite. In figure (ii) both f'(a-) and f'(a+) exist but they are not equal. Hence f'(a) does not exist. Figure (iii) and (iv) have infinite derivatives, i.e. f'(a) = + ∞ and f' (a) = - ∞ respectively. In case of figure (v) we have f'(a) = + ∞ and f'(a+) = - ∞ and hence f'(a) does not exist.

Note:  "Differentiability implies continuity but continuity does not imply differentiability"

Let y = f(x) be continuous in (a, b). Then the derivative or differential coefficient of f(x) w.r.t. x at x ∈ (a, b), denoted by dy/dx or f'(x), is

dy/dx = lim?x→0 (f(x+?x) - f(x)) /?x                            ..... (1)

provided the limit exists and is finite and the function is said to be differentiable.

## To find the derivative of f(x) from the first Principle

If we obtain the derivative of y = f(x) using the formula dy/dx = limh→0 (f(x+h)-f(x))/h, we say that we are finding

For example, y = cos 2x.

Here f(x) = cos 2x and

Right Hand Derivative

Right hand derivative of f(x) at x = a is denoted by, Rf'(a) or f'(a+) and is defined as

Rf'(a)= limh→0 (f(a+h)-f(a))/h,   h>0

Left Hand Derivative

Left hand derivative of f(x) at x = 1 is denoted by Lf'(a) or f'(a-) and is defined as

Lf'(a)= limh→0 (f(a-h)-f(a))/(-h),  h>0

Clearly, f(x) is differentiable at x=a if and only if Rf'(a)=Lf'(a).

Illustration:

Consider the function

f(x) =

Since, the function f(x) changes its value at x = 0, there is a possibility of its being non-differentiable at x = 0

Consider,   f'(0+) = limh→0 (f(0+h)-f(0))/h = limh→0 sin (1/h)

= does not exist

Also, f' (0-) = limh→0 (f(0-h)-f(0))/(-h) = limh→0 -sin( 1/h) (does not exist.)

=> f' (0) does not exist.

Note: limh→0 sin 1/h fluctuates between -1 and 1.

Note: Whenever at x = a, f'(a+) = l(a finite number) and f'(a) = l2 (a finite number) and if l1 ≠ l2, then f' (a) does not exist, but f(x) is a continuous function at x = a.

Illustration:

A function is defined as follows:

f(x)= Discuss the differentiability of the function at x=1.

Solution:

We have R.H.D. = Rf'(1) = limh→0 (f(1-h)-f(1))/h = limh→0 (1+h-1)/h = 1

and L.H.D. = Lf'(1)= limh→0 (f(1-h)-f(1))/(-h)

= limh→0 ((1-h)3-1)/(-h) = limh→0 (3-3h+h2 ) = 3

? Rf'(1)≠ Lf'(1)⇒ f(x) is not differentiable at x=1.

Illustration:

Discuss the continuity and differentiability of the function

f(x)= at x = 0.

Solution:

Since f'(0-) ≠ f' (0+), f(x) is not differentiable at x=0.

## Geometrical Interpretation of Differentiability

Geometrically we interpret f' (x)=limh→0 (f(x+h)-f(x))/h as the slope of the tangent to the graph of y = f(x) at the point (x, f(x)). Thus if there is no tangent line at a certain point, the function is not differentiable at that point. In other words, a function is not differentiable at a corner point of a curve, i.e., a point where the curve suddenly changes direction. See the following graphs:

Note that if a function is discontinuous at the point x = a then three is necessarily a break at that point in the graph of function. So, every differentiable function is continuous but the converse is not true. That is, a continuous function need not be differentiable.

Illustration:

Let: R->R is a function defined by f(x) = max {x, x3}. Find the set of all points on which f(x) is discontinuous and f(x) is not differentiable.

Solution:

We have from the graph of y=x and y=x3

f(x)=

Clearly, there is no break in the graph so f(x) is everywhere continuous but there are sudden change of directions at x=-1, 0, 1. So, f(x) is not differentiable at x ∈ {-1, 0, 1}

## List of Derivatives of Important Functions

• d/dy (xn ) = nx(n-1)

• d/dx (cosec x) = -cosec x cot x

• d/dx (x) = 1

• d/dx (ex ) = ex

• d/dx (1/x) = -1/x2

• d/dx (ln|x| ) = 1/x

• d/dx (1/xn ) = -n/x(n+1) ,   x>0

• d/dx (sin-1 x) = 1/√(1-x2 )

• d/dx (sin x ) = cos x

• d/dx (cos-1 x) = 1/√(1-x2 )

• d/dx (cos x ) = -sin x

• d/dx (tan-1 x) = 1/(1-x2 )

• d/dx (tan x ) = sec2 x

• d/dx (cot-1 x) = (-1)/(1+x2 )

• d/dx (cot x ) = -cosec2 x

• d/dx (sec-1 x) = 1/(|x|√(x2-1))

• d/dx (sec x ) = sec x tan x

• d/dx (cosec-1 x) = (-1)/(|x|√(x2-1))

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