Class 12 Maths Continuity Differentiability Differentiability

Differentiability

A differentiable function of one real variable is a function whose derivative exists at each point in its domain. As a result, the graph of a differentiable function must have a tangent line at each point in its domain. It should be relatively smooth, and cannot contain any breaks, bends, or cusps.

Let f is a real function and c is a point in its domain. The derivative of f at c is defined by

d(f(x))/dx|c = f’(c) = limh->0 {f(c + h) – f(c)}/h

provided this limit exists.

dy/dx is also written as y’ and it read as differentiation of y with respect to x.

If limit limh->0 {f(c + h) – f(c)}/h does not exist, we say that f is not differentiable at c.

In other words, we say that a function f is differentiable at a point c in its domain if both

limh->0- {f(c + h) – f(c)}/h  and limh->0+ {f(c + h) – f(c)}/h are finite and equal.

i.e.  limh->0- {f(c + h) – f(c)}/h  = limh->0+ {f(c + h) – f(c)}/h

A function is said to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b].

 Similarly, a function is said to be differentiable in an interval (a, b) if it is differentiable at every point of (a, b).

Problem: Prove that the function f given by f(x) = |x - 1|, x є R is not differentiable at x = 1.

Solution:

The given function is f(x) = |x - 1|, x є R

It is known that a function f is differentiable at a point x = c in its domain if both

limh->0- {f(c + h) – f(c)}/h and limh->0+ {f(c + h) – f(c)}/h are finite and equal.

To check the differentiability of the given function at x = 1,

Consider the left hand limit of f at x = 1

   limh->0- {f(1 + h) – f(1)}/h

= limh->0- {|1 + h - 1| – |1 - 1|}/h

= limh->0- {|h| – 0}/h

= limh->0- (-h)/h                [Since h < 0 => |h| = -h]

= -1

Consider the right hand limit of f at x = 1

   limh->0+ {f(1 + h) – f(1)}/h

= limh->0+ {|1 + h - 1| – |1 - 1|}/h

= limh->0+ {|h| – 0}/h

= limh->0+ (h)/h                [Since h > 0 => |h| = h]

= 1

Since the left and right hand limits of f at x = 1 are not equal, therefore f is not differentiable at x = 1.

Algebra of Derivatives:

  1. (u ± v)’ = u’ ± v’
  2. (uv)’ = u’v + uv’ [Leibnitz or product rule]
  3. (u/v)’ = (uv’ – uv’)/v2, where v ≠ 0 [Quotient rule]

Theorem: If a function is 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.

Proof: Let f(x) be a function differentiable at x = c. Then

limx->c {f(x) – f(c)}/(x - c) = f’(c)

But for x ≠ c, we have

f(x) – f(c) = [{f(x) – f(c)}/(x - c)] * (x - c)

Now, limx->c [f(x) – f(c)] = limx->c [[{f(x) – f(c)}/(x - c)] * (x - c)]

=> limx->c [f(x)] – limx->c [f(c)] = limx->c [{f(x) – f(c)}/(x - c)] * limx->c (x - c)

=> limx->c [f(x)] – limx->c [f(c)] = f’(c) * (c - c)

=> limx->c [f(x)] – limx->c [f(c)] = f’(c) * 0

=> limx->c [f(x)] – limx->c [f(c)] = 0

=> limx->c [f(x)] = limx->c [f(c)]

=> limx->c [f(x)] = f(c)

Hence, f 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.

For example, the function defined by f(x) = |x| is a continuous function at x = 0 but it is not differentiable at x = 0.

Consider the left hand limit

     limh->0- {f(0 + h) – f(0)}/h = (-h)/h = -1

The right hand limit

limh->0+ {f(0 + h) – f(0)}/h = h/h = 1

Since the above left and right hand limits at 0 are not equal, limh->0 {f(0 + h) – f(0)}/h does not exist and hence f is not differentiable at 0. Thus f is not a differentiable function.

Derivatives of composite functions:

Chain Rule: Let f be a real valued function which is a composite of two functions u and v;

i.e., f = v o u. Let t = u(x) and if both dt/dx and dv/dt exist, we have

df/dx = dv/dt * dt/dx

Again let f = (w o u) o v. If t = v (x) and s = u (t), then

df/dx = d(w o u)/dt * dt/dx =  dw/ds * ds/dt * dt/dx

Problem: Find the derivative of tan (2x + 3).

Solution: Let f(x) = tan (2x + 3), u(x) = 2x + 3 and v(t) = tan t. Then

(v o u) (x) = v(u(x)) = v(2x + 3) = tan (2x + 3) = f(x)

Thus f is a composite of two functions.

Put t = u(x) = 2x + 3

Then dv/dt = sec2 t and dt/dx = 2 exist.

Hence, by chain rule

df/dx = dv/dt * dt/dx = 2 sec2 (2x + 3)

Alternatively,

Let y = tan (2x + 3)

Now, dy/dx = d[tan (2x + 3)]/dx

                      = sec2 (2x + 3) * d(2x + 3)/dx

                      = 2 sec2 (2x + 3)

Derivatives of implicit functions:

When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f(x), we say that y is given as an explicit function of x.

Problem: Find dy/dx of the function: 2x + 3y = sin x

Solution:

Given, 2x + 3y = sin x

Differentiating w.r.t. x, we get

=> d(2x + 3y)/dx = d(sin x)/dx

=> d(2x)/dx + d(3y)/dx = d(sin x)/dx

=> 2 + 3 * dy/dx = cos x

=> 3 * dy/dx = cos x – 2

=> dy/dx = (cos x – 2)/3   

Alternatively,

Given, 2x + 3y = sin x

=> 3y = sin x – 2x

Differentiating w.r.t. x, we get

       d(3y)/dx = d(sin x – 2x)/dx

=> 3 * dy/dx = d(sin x)/dx – d(2x)/dx

=> 3 * dy/dx = cos x – 2

=> dy/dx = (cos x – 2)/3

Derivatives of inverse trigonometric functions:

Problem: Find dy/dx of the function: y = tan-1{(3x – x3)/(1 - 3x2)}, -1/√3 < x < 1/√3

Solution:

Given, y = tan-1{(3x – x3)/(1 - 3x2)}

=> tan y = (3x – x3)/(1 - 3x2)   ……..1

It is know that tan y = (3 * tan y/3 – tan3 y/3)/(1 – 3 tan2 y/3)    …….2

Comparing equation 1 and 2, we get

x = tan y/3         ………….3

Differentiating it w.r.t. x, we get

      d(x)/dx = d(tan y/3)/dx

=> 1 = sec2 y/3 * d(y/3)/dx

=> 1 = sec2 y/3 * (1/3) * dy/dx

=> dy/dx = 3/(sec2 y/3)

=> dy/dx = 3/(1 + tan2 y/3)

=> dy/dx = 3/(1 + x2)               [From equation 3]

 

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