## NCERT Notes for Class 12 Mathematics

## Chapter 5:
Continuity and Differentiability

### Derivative

The rate of change of a quantity y with respect to another quantity x is called the derivative or
differential coefficient of y with respect to x .

### Differentiation of a Function

Let f(x) is a function differentiable in an interval [a, b]. That is, at every point of the interval,
the derivative of the function exists finitely and is unique. Hence, we may define a new
function g: [a, b] ? R, such that, ? x ? [a, b], g(x) = f'(x).

This new function is said to be differentiation (differential coefficient) of the function f(x) with respect to x and it is denoted by df(x) / d(x) or Df(x) or f'(x).

### Differentiation from First Principle

Let f(x) is a function finitely differentiable at every point on the real number line. Then, its
derivative is given by

#### Standard Differentiations

1. d / d(x) (x^{n}) = nx^{n 1}, x ? R, n ? R

2. d / d(x) (k) = 0, where k is constant.

3. d / d(x) (e^{x}) = e^{x}

4. d / d(x) (a^{x}) = a^{x} loge a > 0, a ? 1

#### Fundamental Rules for Differentiation

(v) if d / d(x) f(x) = ?(x), then d / d(x) f(ax + b) = a ?(ax + b)

(vi) Differentiation of a constant function is zero i.e., d / d(x) (c) = 0.

#### Geometrically Meaning of Derivative at a Point

Geometrically derivative of a function at a point x = c is the slope of the tangent to the curve y
= f(x) at the point {c, f(c)}.

Slope of tangent at P = lim _{x ? c} f(x) f(c) / x c = {df(x) / d(x)} _{x = c} or f (c).

##### Different Types of Differentiable Function

###### 1. Differentiation of Composite Function (Chain Rule)

If f and g are differentiable functions in their domain, then fog is also differentiable and

(fog) (x) = f {g(x)} g (x)

More easily, if y = f(u) and u = g(x), then dy / dx = dy / du * du / dx.

If y is a function of u, u is a function of v and v is a function of x. Then,
dy / dx = dy / du * du / dv * dv / dx.

###### 2. Differentiation Using Substitution

In order to find differential coefficients of complicated expression involving inverse
trigonometric functions some substitutions are very helpful, which are listed below .

###### 3. Differentiation of Implicit Functions

If f(x, y) = 0, differentiate with respect to x and collect the terms containing dy / dx at one side
and find dy / dx.

Shortcut for Implicit Functions For Implicit function, put d /dx {f(x, y)} = ?f / ?x / ?f / ?y,
where ?f / ?x is a partial differential of given function with respect to x and ?f / ?y means
Partial differential of given function with respect to y.

###### 4. Differentiation of Parametric Functions

If x = f(t), y = g(t), where t is parameter, then dy / dx = (dy / dt) / (dx / dt) = d / dt g(t) / d / dt f(t) = g (t) / f (t)

###### 5. Differential Coefficient Using Inverse Trigonometrical Substitutions

Sometimes the given function can be deducted with the help of inverse Trigonometrical
substitution and then to find the differential coefficient is very easy.

#### Logarithmic Differentiation Function

(i) If a function is the product and quotient of functions such as y = f_{1}(x) f_{2}(x) f_{3}(x)
/ g_{1}(x)
g_{2}(x) g_{3}(x)
, we first take algorithm and then differentiate.

(ii) If a function is in the form of exponent of a function over another function such as
[f(x)]^{g(x)} , we first take logarithm and then differentiate.

#### Differentiation of a Function with Respect to Another Function

Let y = f(x) and z = g(x), then the differentiation of y with respect to z is dy / dz = dy / dx / dz / dx = f (x) / g (x)

#### Successive Differentiations

If the function y = f(x) be differentiated with respect to x, then the result dy / dx or f (x), so obtained is a function of x (may be a constant).

Hence, dy / dx can again be differentiated with respect of x.

The differential coefficient of dy / dx with respect to x is written as d /dx (dy / dx) = d^{2}y / dx^{2} or f (x). Again, the differential coefficient of d^{2}y / dx^{2} with respect to x is written as d / dx (d^{2}y / dx^{2}) = d^{3}y / dx^{3} or f'(x)

Here, dy / dx, d^{2}y / dx^{2}, d^{3}y / dx^{3},
are respectively known as first, second, third,
order differential coefficients of y with respect to x. These alternatively denoted by f (x), f (x), f

(x),
or y_{1}, y_{2}, y_{3}
., respectively.

Note dy / dx = (dy / d?) / (dx / d?) but d^{2}y / dx^{2} ? (d^{2}y / d?^{2}) / (d^{2}x / d?^{2})

#### Leibnitz Theorem

If u and v are functions of x such that their nth derivative exist, then

#### nth Derivative of Some Functions

#### Derivatives of Special Types of Functions

##### (vii) Differentiation of a Determinant

##### (viii) Differentiation of Integrable Functions

If g_{1} (x) and g_{2 }(x) are defined in [a, b], Differentiable at x ? [a, b] and f(t) is continuous for g_{1}(a) ? f(t) ? g_{2}(b), then

##### Partial Differentiation

The partial differential coefficient of f(x, y) with respect to x is the ordinary differential coefficient of f(x, y) when y is regarded as a constant. It is a written as ?f / ?x or D_{x}f or f_{x}.

e.g., If z = f(x, y) = x_{4} + y^{4} + 3xy^{2} + x^{4}y + x + 2y

Then, ?z / ?x or ?f / ?x or f_{x} = 4x^{3} + 3y^{2} + 2xy + 1 (here, y is consider as constant) ?z / ?y or ?f / ?y or fy = 4y^{3} + 6xy + x^{2} + 2 (here, x is consider as constant)

##### Higher Partial Derivatives

Let f(x, y) be a function of two variables such that ?f / ?x , ?f / ?y both exist.

(i) The partial derivative of ?f / ?y w.r.t. x is denoted by ?^{2}f / ?x^{2} / or fxx.

(ii) The partial derivative of ?f / ?y w.r.t. y is denoted by ?^{2}f / ?y^{2} / or fyy.

(iii) The partial derivative of ?f / ?x w.r.t. y is denoted by ?^{2}f / ?y ?x / or fxy.

(iv) The partial derivative of ?f / ?x w.r.t. x is denoted by ?^{2}f / ?y ?x / or fyx.

Note ?^{2}f / ?x ?y = ?^{2}f / ?y ?x

These four are second order partial derivatives.

##### Eulers Theorem on Homogeneous Function

If f(x, y) be a homogeneous function in x, y of degree n, then x (&partf / ?x) + y (&partf / ?y) = nf

##### Deduction Form of Eulers Theorem

If f(x, y) is a homogeneous function in x, y of degree n, then

(i) x (?^{2}f / ?x^{2}) + y (?^{2}f / ?x ?y) = (n 1) &partf / ?x

(ii) x (?^{2}f / ?y ?x) + y (?^{2}f / ?y^{2}) = (n 1) &partf / ?y

(iii) x^{2} (?^{2}f / ?x^{2}) + 2xy (?^{2}f / ?x ?y) + y^{2} (?^{2}f / ?y^{2}) = n(n 1) f(x, y)

##### Important Points to be Remembered

If ? is m times repeated root of the equation f(x) = 0, then f(x) can be written as f(x) =(x ?)^{m} g(x), where g(?) ? 0.

From the above equation, we can see that f(?) = 0, f (?) = 0, f (?) = 0,
, f^{(m l)} ,(?) = 0.

Hence, we have the following proposition f(?) = 0, f (?) = 0, f (?) = 0,
, f^{(m l)} ,(?) = 0.

Therefore, ? is m times repeated root of the equation f(x) = 0.

**PART 2**