**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 (orx) 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) = x ^{2}**

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 x^{2} + 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

Ex. Find dy/dx | Step 1 Let y=u/v where, u=x^{2}, v=(x+2) | Step 2 |

Step 3 = 1 + 0 = 1 | Step 4 | Step 5 | Step 6 |

Step 7 | Step 8 |

**Derivatives of Mathematical Functions – Problems**

**Derivative of Functions – Practical Problems and Solutions**

Ex. 1 : y = x^{4} + 3x | ||||

Step 1 : | Step 2 : | Step 3 : | ||

Ex. 2: y = 3x^{5} – 5x^{3} + 20 | ||||

Step 1 : | Step 2 : | |||

Step 3 : | Step 4 : = 3.5x^{5-1 }– 5.3 x^{3-1 } = 15x^{4 }– 15x^{2}. | |||

Ex. 3: y = – 8 + (4/5)x^{5} – (7/8)x^{8} | ||||

Step 1 : | Step 2 : | |||

Step 3 : ^{}^{ }= 4x^{4 }– 7x^{7}^{} | ||||

**Derivative of Product of two functions**

**Derivative of Product of functions**

**Ex.1**

y = (x + 2)(2x^{3} – 21)

**Step 1**

Letu = x + 2, v = 2x^{3 }– 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**

= 6x^{3} + 12x^{2} + 2x^{3} – 10

= 8x^{3} + 12x^{2} – 10.

**Ex. y = x(x ^{2} – 2)(x^{3} + 3)**

**Step 1 : **u = x , v = x^{2} – 2, w = x^{3} + 3

**Step 2 : **

**Step 3**

**Step 4**

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

= (x^{5} – 2x^{3} + 3x^{2 }– 6) + (2x^{5 }+ 6x^{2}) + (3x^{5 }– 6x^{3})

= (x^{5 }+ 2x^{5 }+ 3x^{5}) – (2x^{3} + 6x^{3}) + (3x^{2 }+ 6x^{2}) – 6

= 6x^{5} – 8x^{3} + 9x^{2 }– 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**

Step 1 Letu = (x + 1)(2x^{2} – 1) = 2x^{3} + 2x^{2} – x – 1, v = x^{2} + 1 | Step 2 , |

Step 3 | Step 4 |

**Ex. **

**Step 1 : **Letu = (x + 1)(2x^{2} – 1) = 2x^{3} + 2x^{2} – x – 1, v = x^{2} + 1

**Step 2 : **,

**Step 3 : **** **

**Step 4 : **

**Derivative of Logarithmic functions**

**Derivative of logarithmic functions**

Ex. 1: y = 6x^{2} + 4x − 25 | Step 1 : dy/dx= | Step 2: = (6 X 2x) +( 4.1) − 0 = 12x + 4 |

Ex. 2: h(x)=a^{x} + x^{b} + a^{b}, where y is a function of x, a & b are constant | Step 1 Step 2 |

**Ex. 1:** **y = 6x ^{2} + 4x – 25**

**Step 1:** dy/dx

**Step 2: **= (6 X 2x) + ( 4.1) − 0 = 12x + 4

**Ex. 2: h(x) = a ^{x} + x^{b} + a^{b}, where y is a function of x, a & b are constant**

**Step 1 : **

**Step 2 : **=

Ex. 3: f(x)= | Step 1: | Step 2: |

**Ex. 3: f(x) = **

**Step 1: **** **

**Step 2:**

**Ex. 4 : **f(x) = e^{x} log x

**Step 1: **

**Step 2:**

**Step 3: **** **

**Derivative of Exponential functions **

**Derivative of Exponential functions**

**Ex.1 :** y = 3^{x} . x^{8}

**Step 1 :** dy/dx ** **

**Step 2 : **= (x^{8} 3^{x} log_{e} 3) + (7.3^{x}.x^{7}) = x^{8} 3^{x} log_{e} 3 + 7.3^{x}.x^{7}

**Ex.2 : f(x) = x ^{2}/e^{x}**

**Step 1 : **** **

**Step 2 : ** = [2xe^{x }– x^{2}e^{x }= e^{x}(2x – x^{2})] e^{x }[x(2 – x)]

**Step 3 : **** **

**Ex.3 :** **y =**** e ^{x }/ log x**

**Step 1 : **

**Step 2 : **** **

**Step 3 : **

**Step 4 : **

**Ex.4 : ****h(x) = 5 ^{x }. log(x)**

**Step 1 : **Let f(x) = 5^{x }, 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 = z^{2} + 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 = 2z^{2} + 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 x^{2} + xy – y^{2} = 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, y
^{2}, will be differentiated as 2y ´ {(dy/dx)} and 3y^{5}as 15y^{4}´ {(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. x^{2} – y^{2} + 5x = 7y Find (dy/dx) | Step 1: Differentiate both sides w.r.t. x, | Step 2: |

Step 3: | Step 4 : By transposing | Step 5: |

**Ex.** x^{2} – y^{2} + 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 = x ^{x}. 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 = x ^{x} . e^{2(x+3}**

**Step 1 : **Let u = x^{x }, v = e^{2(x + 3)}. So, y = u.v

**Step 2 : **

**Step 3: **u = x^{x}, so, log u = x log x. Hence,

**Step 4 : **

**Step 5 : **= x^{x} (log x + 1)

**Step 6: **

**Step 7 : ** = 2.e^{2(x + 3)}

**Step 8 : **** **

**Step 9 : **= x^{x}. e^{2(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 (d^{2}y/dx^{2}), d^{3}y/dx^{3} …etc

**Ex.1: y = 4x ^{3 }– 10x. **Find 2

^{nd}order differential co-efficient w. r. t. x

**Step 1: **= 12x^{2} – 10

**Step 2 :** 2^{nd} order differentiation,

**Ex.2: **Differentiate e^{2x}. with respect to e^{x}. Let y = e^{2x}

**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£ x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6} £ b, then f(x_{6}) > f(x_{5}) …> f(x_{2}) > f(x_{1}).

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£ x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6} £ b, then f(x_{6}) < f(x_{5}) …< f(x_{2}) <f(x_{1}).

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 = x ^{4} – x^{2}**. 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)x Show that y increases for all increase value of x^{3}-3x^{2}+9x. | Step 1: =(x-3)^{2} | So, Which is positive for al values of x |

**Ex. 2 : y = (1/3)x ^{3 }– 3x^{2 }+ 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

Ex. 2 : y = (1/3)x Show that y increases for all increase value of x^{3 }– 3x^{2 }+ 9x. | Step 1: = (x–3)^{2} | So, Which is positive for al values of x |

**Maximum and Minimum value of a function**

**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 = (x^{2}/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**

Ex.1: y = x^{3} – 9x ^{2} + 24 x +12 Find the max & min values of y | Step 1: | Step 2: for Stationary values, dy/dx=0 So,3x ^{2} – 18 x + 24 = 0 |

Step 3 x^{2}-6x+8=0 | Step 4 x^{2}-4x-2x+8=0 | Step 5 x(x-4)-2(x-4)=0 | Step 6 (x-4)(x-2)= 0 | Step 7 x-4=0,i.e. x=4 x-2=0,i.e. x=2 |

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

**Ex.1 : y = x ^{3} – 9x^{2} + 24 x + 12. Find the maximum & minimum values of y**

**Step 1 : **

**Step 2 : **for Stationary values, dy / dx = 0, So, 3x^{2} – 18x + 24 = 0

**Step 3 : **So,x^{2 }– 6x + 8 = 0

**Step 4 : **or,x^{2 }– 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 = x^{3 }– 9x^{2 }+ 24x + 12.

Putting x = 2, y_{max }= (2)^{3}– 9.(2)^{2 }+ 24.(2) + 12= 8 – 36 + 48 + 12 = 32

putting x = 4, y_{min }= (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 – (x^{2}/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

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