Differential Calculus

## Differential Calculus

Calculus is the branch of mathematics of computing derivatives and integrals of functions, that studies the rates at which quantities change, based on methods of summation of infinitesimal differences.

Differential Calculus and Integral Calculus are the two main components of mathematics of Calculus

• Differential Calculus  is the branch of mathematics concerned chiefly with the study of the rate of change of functions with respect to their variables especially through the use of derivatives and differentials.
• Integral Calculus is the branch of mathematics that deals with integrals, especially the methods of  ascertaining indefinite integrals and applying them to the solution of differential equations

Differentiation

Differentiation is the process of finding the derivative of a continuous function.

Derivative

Derivative is defined as the limiting value of the ratio of the change (increment) in the function corresponding to a small change (increment) in the independent variable (argument), as it tends to zero.

## Derivative of a mathematical function

Let y = f(x) be a function. If h (or x) be the small increment in x and the corresponding increment in y or f(x) be y = f(x + h) − f(x) then the derivative of function f(x) is defined as

The derivative of a mathematical function f(x) is also known as differential co efficient of f(x) with respect to x and is denoted as f’(x)

A function f(x) is said to be differentiable at x = c, if exists at x = c. It is denoted by f´(c) or (dy/dx)x=c

Ex. Find the derivative of f(x) = x2

Thus derivative of f(x) exists for all values of x and equals 2x at any point x.

## Derivatives of Mathematical Function – Problems

Derivative of Simple Function- Problem and Solution

Example: If first principle find Let Now Example:  If y = 3, find from first principle.

f(x) = 3, then f(x + h) = 3, f(x + h) – f(x) = 3 – 3 = 0

(In general c = 0, where c constant)

## Derivatives of Mathematical Function – Problems

Derivative of Discontinuity | Derivative of Constant

Example 1: Find from first principle, the derivative of at x = 0, 2.

Let At x = 0, f(x) has no existence (as is undefined) at x = 0, f´(x) also has no existence

Ex.2 Find the derivative of x2 + 4x.

Derivatives of Constant

Ex. where c is a constant.

## Differentiation Standard Formulae

Standard Formulae of Differentiation

### Derivative of the algebraic sum of two derivables

The derivative of the algebraic sum (or difference) of two derivable function is the algebraic sum (or difference) of their derivatives.

### Derivative of product of two functions

The derivative of product of two functions = first function x derivative of the second function + second function x derivative of the first function.

## Derivative of Product of Function & Constant

The derivative of product of a function and a constant is the product of the constant and the derivative of the function

### Derivative of Quotient of two functions

Derivative of the quotient of two derivable functions = [(denominator x derivative of numerator –numerator x derivative of denominator) / (denominator)2, where denominator ¹ 0

## Derivatives of Mathematical Functions – Problems

Derivative of Functions – Practical Problems and Solutions

## Derivative of Product of two functions

Derivative of Product of functions

Ex.1

y = (x + 2)(2x3 – 21)

Step 1

Letu = x + 2, v = 2x3 – 21, So, y = u.v

Step 2 =1 + 0 =1

Step 3

Step 4 : substituting value of u, v, du/dx,  dv/dx

Step 5

= 6x3 + 12x2 + 2x3 – 10

= 8x3 + 12x2 – 10.

Ex. y = x(x2 – 2)(x3 + 3)

Step 1 : u = x , v = x2 – 2, w = x3 + 3

Step 2 :

Step 3

Step 4

= [{(x2 – 2) (x3 + 3).1} + {x(x3 + 3).2x} + {x (x2 – 2).3x2}]

= (x5 – 2x3 + 3x2 – 6) + (2x5 + 6x2) + (3x5 – 6x3)

= (x5 + 2x5 + 3x5) – (2x3 + 6x3) + (3x2 + 6x2) – 6

= 6x5 – 8x3 + 9x2 – 6

## Derivative of Quotient of two functions

Derivative of Quotient of functions

Ex.

Step 1: Let u = x – 2, v = x + 2

Step 2 :

Step 3 : Step 4 :

Step 5 :

## Derivative of Product & Quotient of functions

Derivative of Product & Quotient of functions

Ex.

Step 1 : Letu = (x + 1)(2x2 – 1) = 2x3 + 2x2 – x – 1, v = x2 + 1

Step 2 : , Step 3 :

Step 4 :

## Derivative of  Logarithmic functions

Derivative of  logarithmic functions

Ex. 1: y = 6x2 + 4x – 25

Step 1: dy/dx Step 2: = (6 X 2x) + ( 4.1) − 0 = 12x + 4

Ex. 2: h(x) = ax + xb + ab, where y is a function of x, a & b are constant

Step 1 : Step 2 : =

Ex. 3: f(x) =

Step 1:

Step 2:

Ex. 4 : f(x) = ex log x

Step 1:

Step 2:

Step 3:

## Derivative of  Exponential functions

Derivative of  Exponential functions

Ex.1 : y = 3x . x8

Step 1 : dy/dx

Step 2 : = (x8 3x loge 3) + (7.3x.x7) = x8 3x loge 3 + 7.3x.x7

Ex.2 : f(x) = x2/ex

Step 1 :

Step 2 : = [2xex – x2ex = ex(2x – x2)] ex [x(2 – x)]

Step 3 :

Ex.3 : y = ex / log x

Step 1 :

Step 2 :

Step 3 :

Step 4 :

Ex.4 : h(x) = 5x . log(x)

Step 1 : Let f(x) = 5x , g(x) = log(x)

Step 2 :

Step 3 :

Step 4 :

Step 5 :

## Derivative of a Function of a Function

Derivative of Function of a Function

A variable y may be a function of a second variable z, which again may be function of a third variable x

Ex. y = z2 + 3 and z = 2x + 1. Here y is a function of z and z is again a function of x. So,Y depends on x. So, here y is called the function of another function

Ex. y = 2z2 + 1, z = 4x – 2, Find derivative of y with respect to x

Step 1: , Step 2: Step 3: ## Differentiation of Implicit Functions

Derivative of Implicit Function

An implicit function is a function that is defined implicitly by an implicit equation, by associating one of the variables (the value) with the others (the arguments). In simple words, Implicit function means a function y of variable x, where y can not be expressed in terms of x only. For example, the equation x2 + xy – y2 = 1 represents an implicit relation.

Steps of Differentiation of Implicit Function

• Differentiate both the sides of the given relation with respect to x. While differentiating the y terms not containing x, multiply the same coefficient by dy/dx (so, y2, will be differentiated as 2y ´ {(dy/dx)} and 3y5 as 15y4 ´ {(dy/dx)}
• Transfer all the terms containing (dy/dx) to one side and all the other terms on another side of the equation
• Divide both the sides by co-efficient of (dy/dx) and thereby get (dy/dx)
• Simplify the result using the relation, to the extent possible

Differentiation of Implicit Functions

Ex. x2 – y2 + 5x = 7y. Find (dy/dx)

Step 1: Differentiate both sides w. r. t. x, we get,

Step 2:

Step 3 :

Step 4 : By transposing,

Step 5 :

## Differentiation of Logarithmic Functions

The process of finding out derivative by taking logarithm in the first instance is termed as logarithmic differentiation. The procedure is easy to adopt when the function to be differentiated involves a function in its power or when the function is the product of number of functions.

Ex : y = xx. Find dy/dx

Step 1 : Taking log of both sides, we get, log y = x log x

Step 2 : Taking derivative of both sides, we get,

Step 3 :

## Differentiation of Logarithmic Functions – Problems

Logarithmic Differentiation – Problems & Solutions

Ex. y = xx . e2(x+3

Step 1 : Let u = xx , v =  e2(x + 3). So, y = u.v

Step 2 : Step 3: u = xx,  so, log u = x log x. Hence, Step 4 : Step 5 : = xx (log x + 1)

Step 6: Step 7 : = 2.e2(x + 3)

Step 8 :

Step 9 : = xx. e2(x + 3)  ( log x + 3)

## Differentiation of functions of higher order

As derivative of a function is also a function of x, it is possible to carry out successive differentiation of higher order, like (d2y/dx2), d3y/dx3 …etc

Ex.1: y = 4x3 – 10x. Find 2nd order differential co-efficient w. r. t. x

Step 1: = 12x2 – 10

Step 2 : 2nd order differentiation, Ex.2: Differentiate e2x. with respect to ex. Let y = e2x

Step 1: ## Increasing and Decreasing Functions

Increasing & Decreasing function & Stationery Point

Increasing function :  y = ¦(x) is said to be an increasing Function of x, if ¦(x) increases for increasing value of x in a certain interval [a, b].

In an increasing function in interval [ a, b ], say a£ x1 < x2 < x3 < x4 < x5 < x6 £ b, then f(x6) > f(x5) …> f(x2) > f(x1).

In this case the tangent at any point (within the interval) makes an acute angle with the

positive direction of x axis, i.e., dy/dx>0 and the curve y=f(x) rising at that interval

Decreasing function: y = ¦(x) is said to be an decreasing Function of x, if ¦(x) decreases for increasing value of x in a certain interval [a, b].

In an decreasing function in interval [ a, b ], say a£ x1 < x2 < x3 < x4 < x5 < x6 £ b, then f(x6) < f(x5) …< f(x2) <f(x1).

In such case, the tangent at any point makes an obtuse angle with the positive direction of x-axis, i.e dy/dx <0 and the curve of y = f(x) will be falling at that interval

Stationary Point : A function y = f(x) may be neither increasing nor decreasing at some point, then that point is called as stationary point. At stationary point the derivative is zero, i.e dy/dx=0. At the stationary point (S) the curve will be neither rising nor falling.

Ex. 1 : y = x4 – x2. Examine, whether y is increasing or decreasing at x = – 1

Step 1 :

Step 2 : For x = –1,

So, y is a decreasing function, and the tangent at x = – 1 makes an obtuse angle with positive direction of x-axis.

Ex. 2 : y = (1/3)x3 – 3x2 + 9x. Show that y increases for all increase value of x

Step 1: = (x–3)2 .So, , which is positive for al values of x

## Maximum and Minimum value of a function

Maxima and Minima

18.1 Maximum Value of a function : A function is said to be maximum at x = a if its value is maximum in the immediate neighbourhood of x = a, i.e. f (a) > f (a – h), i.e. f (a) > f (a + h), where h is a small positive number

In such case, f’¢(a)=0, f¢¢(a)<0

18.2 Minimum Value of a function : If the value of a function is minimum in the immediate neighbourhoood of a, the function is said to be minimum at x = a, i.e.  f (a) < f (a – h), i.e.  f (a) < f (a + h), where h is a small positive number. In such case, f’¢(a)=0, f¢¢(a)>0

18.3 Steps of obtaining maximum or minimum values of a function

• Find the first derivative dy/dx of the function.
• Putting dy/dx = 0, solve the equation and obtain the values of x. These values of x give stationary points.
• To decide whether these stationary points give maximum values or minimum values, find second derivative. Put these values of x alternately in the second derivative.
• The value of x for which the second derivative is negative gives maximum value while the value for which the second derivative is positive gives minimum value.
• The maximum and minimum values of a function are obtained by putting these values of x in the given function.

Ex: The cost function of a commodity is C = (x2/20) + 30x+150, where x is the number of units produced. Find marginal cost when 60 units are produced.

Step 1 : Marginal Cost =

Step 2 : Putting x = 60, = 6 + 30 = 36

## Maximum and Minimum value – Problems and Solutions

Maxima and Minima – Problems and Solution

At  x = 4  and x =2 we get stationary values

Ex.1 : y = x3 – 9x2 + 24 x + 12. Find the maximum  & minimum values of y

Step 1 : Step 2 : for Stationary values, dy / dx = 0, So, 3x2 – 18x + 24 = 0

Step 3 :  So,x2 – 6x + 8 = 0

Step 4 : or,x2 –  4x – 2x + 8 = 0

Step 5 : or, x(x – 4) – 2(x – 4) = 0

Step 6 : or, (x – 4)(x – 2)= 0

Step 7 : So, x – 4 = 0, i.e. x = 4 & x – 2 = 0, i.e. x = 2

At  x = 4  and x =2 we get stationary values

Maxima & Minima Values

Step 8 : Step 9 : = – 6 (negative). So y is max at x = 2

Step 10 = 6 (positive).  So y is min at x = 4

Step 11 : y = x3 – 9x2 + 24x + 12.

Putting x = 2, ymax = (2)3– 9.(2)2 + 24.(2) + 12= 8 – 36 + 48 + 12 = 32

putting x = 4, ymin = (4)3 – 9.(4)2 + 24.(4) + 12 = 64 – 144 + 96 + 12 = 28

Ex.2 : The cost function of a factory is C = 29 + 15x – (x2/5). Find marginal cost. Also find the units produced when the marginal cost is 14

Step 1 : Marginal Cost  =

Step 2 : At Marginal Cost (i.e.  when dc / dx = 14), 15 – (2x/5) = 14, or 2x/5 = (15 –14) = 1, x = 5/2

So, when Marginal Cost is 14, units produced is 5/2 units or 2½ units