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) = lim_{h->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 lim_{h->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

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

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

lim_{h->0}- {f(c + h) – f(c)}/h and lim_{h->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

lim_{h->0}- {f(1 + h) – f(1)}/h

= lim_{h->0}- {|1 + h - 1| – |1 - 1|}/h

= lim_{h->0}- {|h| – 0}/h

= lim_{h->0}- (-h)/h [Since h < 0 => |h| = -h]

= -1

Consider the right hand limit of f at x = 1

lim_{h->0}+ {f(1 + h) – f(1)}/h

= lim_{h->0}+ {|1 + h - 1| – |1 - 1|}/h

= lim_{h->0}+ {|h| – 0}/h

= lim_{h->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:**

- (u ± v)’ = u’ ± v’
- (uv)’ = u’v + uv’ [Leibnitz or product rule]
- (u/v)’ = (uv’ – uv’)/v
^{2}, 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

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

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

=> lim_{x->c} [f(x)] – lim_{x->c} [f(c)] = f’(c) * (c - c)

=> lim_{x->c} [f(x)] – lim_{x->c} [f(c)] = f’(c) * 0

=> lim_{x->c} [f(x)] – lim_{x->c} [f(c)] = 0

=> lim_{x->c} [f(x)] = lim_{x->c} [f(c)]

=> lim_{x->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

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

The right hand limit

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

Since the above left and right hand limits at 0 are not equal, lim_{h->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 = sec^{2} t and dt/dx = 2 exist.

Hence, by chain rule

df/dx = dv/dt * dt/dx = 2 sec^{2} (2x + 3)

Alternatively,

Let y = tan (2x + 3)

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

= sec^{2} (2x + 3) * d(2x + 3)/dx

= 2 sec^{2} (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 – x^{3})/(1 - 3x^{2})}, -1/√3 < x < 1/√3**

**Solution:**

Given, y = tan^{-1}{(3x – x^{3})/(1 - 3x^{2})}

=> tan y = (3x – x^{3})/(1 - 3x^{2}) ……..1

It is know that tan y = (3 * tan y/3 – tan^{3} y/3)/(1 – 3 tan^{2} 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 = sec^{2} y/3 * d(y/3)/dx

=> 1 = sec^{2} y/3 * (1/3) * dy/dx

=> dy/dx = 3/(sec^{2} y/3)

=> dy/dx = 3/(1 + tan^{2} y/3)

=> dy/dx = 3/(1 + x^{2}) [From equation 3]

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