User talk:Mani
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 = nxn-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*x8-1 = 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.x4-1 + 2.2.x2-1 -10.x1-1 = 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 = [2nd function * derivati of the 1st function - 1st function * derivative of 2nd 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)
== MATHEMATICS FOR PHYSICS