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MATHEMATICS FOR PHYSICS

                                        Differentiation:

Constant quantity:

If the value of a quantity remains the same in a mathematical operation, it is called a constant quantity.

Examples: Integers (4, 7, 13, ...), the fraction (1/2, 4/5), π, e, etc.,

Variable quantity:

If the quantity takes different values in a mathematical operation, it is a variable quantity. Examples: (i) In the equation y = 2x + 3, x and y are variables. (ii) In the equation F = ma where F, m and a are variables.

Variable quantities are divided into two types

(a) Independent quantity.
(b) Dependent quantity. 

In the equation y = 2x + 3, x is the independent quantity, and y is the dependent quantity.

Function:

If corresponding to any given value of x, there exists a single definite value then y is called a function of x. This is represented as y = f(x). This equation means that corresponding to one value of x, there is a single definite value of the variable y. Example: (i) Let y = 2x, when x = 1 then y = 2; when x = 2 then y = 4... Thus corresponding to each value of x, there is a definite value of y.

Difference and Differential coefficient:

Let y be a function of x. This is denoted as y = f(x). Here x and y are the variables. Let the value of x change to x+△x and correspondingly the value of y changes to y+△y.

The change between the initial and final values of a variable quantity is called Difference.

Difference in x = (x+ △x) - x = △x; Difference in y =(y+△y)-y=△y

The ratio of △y/△x is called the quotient of the two increments. When Difference in x (i.e., △x) is very small i.e., almost approaching to zero, then

we write △y/△x is equal to dy/dx.

In mathematical language, we represent the above statement as:

lim(∆x→0)⁡█(∆y/∆x) =dy/dx or

Lt(∆x→0)⁡█(∆y/∆x) =dy/dx

In this equation, ∆y/∆x is the ratio of a small quantity ∆y to another small quantity ∆x. But dy/dx is a single quantity and is called the differential coefficient of y with respect to x.

dy/ dx = Lt(∆x→0)⁡█(∆y/∆x) = Lt(∆x→0)⁡█[f(x+∆x)-f(x)]/∆x

The rate of change of a dependent variable with respect to the independent variable is called the differential coefficient or derivative.

d/dx does not mean that d is divided by dx. It is a single operator called the differential operator.

Differentiation:

The process of finding the differential coefficient of a function is called differentiation.

Basic Theorems on Differentiation: i) The derivative of a constant is zero. Let y = f(x) = c. Where c is constant.

Then dy/dx =d/dx(c) =0

Example: Differentiate y = 2a where a is a constant dy/dx= d/dx(2a) =0

ii) The differential coefficient of xⁿ is obtained by decreasing the power of x by unity and multiplying by n.

If y = xⁿ then dy/dx = nx^n-1, where n may be positive or negative.

Example: Differentiate y = 8x⁸ w.r.t. x.

Solution: dy/dx = d/dx(8x⁸) = 8d/dx(x⁸) = 8*8*x^8-1 = 64x⁷

iii) The derivative of the algebraic sum of two functions is equal to the algebraic sum of the derivatives of the two functions.

Let y = u±v±w±..... where u, v, w..... are all functions of x.

Then dy/dx = d/dx(u±v±w±.....) = d/dx(u)±d/dx(v)±d/dx(w)± ...

Example: Differentiate y = 3x⁴ + 2x² - 10x w.r.t. x.

Solution:

dy/dx = d/dx(3x4 + 2x2 -10x) = d/dx(3x4) + d/dx(2x2)- d/dx (10x)

= 4.3.x^4-1 + 2.2.x^2-1 -10.x^1-1 = 12x³ + 4x - 10

Rules of differentiation

Product Rule: The differential coefficient of the products of two functions = 1st function x differential coefficient of the 2nd function + 2nd function x differential coefficient 1st function.

Let y = uv where u and v are functions of x. Then dy/dx= d/dx(uv) = u(dv/dx) + v(du/dx)

Example: Differentiate y = x(x² – 2x) w.r.t. x.

Solution: Here u = x, v = x² - 2x

dy/dx = d/dx[x(x²-2x)]=x(d/dx)(x² – 2x) + (x²–2x)d/dx(x). = x(2x-2)+(x² - 2x)*1= 3x² - 4x

Quotient Rule:

The differential coefficient of the quotient of two functions = [2^nd function * derivati of the 1^st function - 1^st function * derivative of 2^nd function] divided by the square of the second function. Let y = u/v where u and v are two functions of x. Then, dy/dx= d/dx[(x² + 1)/(x-1)] w.r.t.x.

Example: Differentiate y = (x2 + 1)/(x - 1) w.r.t. x. Solution: Here u = x2 + 1, v = x - 1. range . [(x2 +1)/(x-1)7=(x-1) dx (x +1)-(x2 +1) & (x - 1) (x-1)2 (x-1)× 2x -(x +1)x1 x? - 2x - 1 (x-1) (x-1)