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| MATHEMATICS FOR PHYSICS
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| Differentiation:
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| <big>Constant quantity:</big>
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| If the value of a quantity remains the same in a mathematical operation, it is called a constant quantity.
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| Examples: Integers (4, 7, 13, ...), the fraction (1/2, 4/5), π, e, etc.,
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| <big>Variable quantity:</big>
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| If the quantity takes different values in a mathematical operation, it is a variable quantity. Examples:
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| (i) In the equation y = 2x + 3, x and y are variables.
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| (ii) In the equation F = ma where F, m and a are variables.
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| Variable quantities are divided into two types
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| (a) Independent quantity.
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| (b) Dependent quantity.
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| In the equation y = 2x + 3, x is the independent quantity, and y is the dependent quantity.
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| <big>Function:</big>
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| 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.
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| Example: (i) Let y = 2x, when x = 1 then y = 2; when x = 2 then y = 4...
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| Thus corresponding to each value of x, there is a definite value of y.
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| <big>Difference and Differential coefficient:</big>
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| 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.
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| The change between the initial and final values of a variable quantity is called Difference.
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| Difference in x = (x+ △x) - x = △x; Difference in y =(y+△y)-y=△y
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| The ratio of △y/△x is called the quotient of the two increments.
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| When Difference in x (i.e., △x) is very small i.e., almost approaching to zero, then
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| we write △y/△x is equal to dy/dx.
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| In mathematical language, we represent the above statement as:
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| lim(∆x→0)█(∆y/∆x) =dy/dx or
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| Lt(∆x→0)█(∆y/∆x) =dy/dx
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| In this equation, ∆y/∆x is the ratio of a small quantity ∆y to another small quantity ∆x.
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| But dy/dx is a single quantity and is called the differential coefficient of y with respect to x.
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| dy/ dx = Lt(∆x→0)█(∆y/∆x) = Lt(∆x→0)█[f(x+∆x)-f(x)]/∆x
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| The rate of change of a dependent variable with respect to the independent variable is called the differential coefficient or derivative.
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| d/dx does not mean that d is divided by dx. It is a single operator called the differential operator.
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| <big>Differentiation:</big>
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| The process of finding the differential coefficient of a function is called differentiation.
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| <big>Basic Theorems on Differentiation:
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| </big>
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| i) The derivative of a constant is zero.
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| Let y = f(x) = c. Where c is constant.
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| Then dy/dx =d/dx(c) =0
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| Example: Differentiate y = 2a where a is a constant dy/dx= d/dx(2a) =0
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| ii) The differential coefficient of x<sup>n</sup> is obtained by decreasing the power of x by unity and multiplying by n.
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| If y = x<sup>n</sup> then dy/dx = nx<sup>n-1</sup>, where n may be positive or negative.
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| Example: Differentiate y = 8x<sup>8</sup> w.r.t. x.
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| Solution: dy/dx = d/dx(8x<sup>8</sup>) = 8d/dx(x<sup>8</sup>) = 8*8*x<sup>8-1</sup> = 64x<sup>7</sup>
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| iii) The derivative of the algebraic sum of two functions is equal to the algebraic sum of the derivatives of the two functions.
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| Let y = u±v±w±.....
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| where u, v, w..... are all functions of x.
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| Then dy/dx = d/dx(u±v±w±.....) = d/dx(u)±d/dx(v)±d/dx(w)± ...
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| Example: Differentiate y = 3x<sup>4</sup> + 2x<sup>2</sup> - 10x w.r.t. x.
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| Solution:
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| dy/dx = d/dx(3x<sup>4</sup> + 2x<sup>2</sup> -10x) = d/dx(3x<sup>4</sup>) + d/dx(2x<sup>2</sup>)- d/dx (10x)
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| = 4.3.x<sup>4-1</sup> + 2.2.x<sup>2-1</sup> -10.x<sup>1-1</sup> = 12x<sup>3</sup> + 4x - 10
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| <big>Rules of differentiation</big>
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| Product Rule:
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| 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.
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| Let y = uv where u and v are functions of x.
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| Then
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| dy/dx= d/dx(uv) = u(dv/dx) + v(du/dx)
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| Example: Differentiate y = x(x<sup>2</sup> – 2x) w.r.t. x.
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| Solution: Here u = x, v = x<sup>2</sup> - 2x
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| dy/dx = d/dx[x(x<sup>2</sup>-2x)]=x(d/dx)(x<sup>2</sup> – 2x) + (x<sup>2</sup>–2x)d/dx(x).
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| = x(2x-2)+(x<sup>2</sup> - 2x)*1= 3x<sup>2</sup> - 4x
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| <big>Quotient Rule:</big>
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| The differential coefficient of the quotient of two functions = [2<sup>nd</sup> function * derivati of the 1<sup>st</sup> function - 1<sup>st</sup> function * derivative of 2<sup>nd</sup> function] divided by the square of the second function.
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| Let y = u/v where u and v are two functions of x.
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| Then, dy/dx= d/dx[(x<sup>2</sup> + 1)/(x-1)] w.r.t.x.
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| Example: Differentiate y = (x2 + 1)/(x - 1) w.r.t. x. Solution: Here u = x2 + 1, v = x - 1.
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| range . [(x2 +1)/(x-1)7=(x-1) dx (x +1)-(x2 +1) & (x - 1)
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| (x-1)2
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| (x-1)× 2x -(x +1)x1 x? - 2x - 1
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| (x-1)
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| (x-1)
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