# Differentiation Assignment Help

## Differentiation Introduction

Differentiation is all about finding rates of change of one quantity compared to another. We need differentiation when the rate of change is not constant. The slope of a curve at a point tells us the rate of change of the quantity at that point. The process of finding a derivative is called differentiation.

## Rules for Differentiation

The Constant Rule

If y = c where c is a constant,

dy/dx = 0

e.g. y = 10 then

dy/dx = 0

dy/dx = 0

The Linear Function Rule

If y = a + bx

dy/dx = b

e.g. y = 10 + 6x then

dy/dx = 6

The Power Function Rule

If y = ax^{n}, a & n are constants

dy/dx = n . a . x^{n-1}

i) y = 4x =>

dy/dx = 4 x ^{0} = 4

ii) y = 4x^{2} =>

dy/dx = 8 x

iii) y = 4x^{3} =>

dy/dx = 12 x^{2}

iv) y = 4x-2 =>

dy/dx = - 8 x^{ - 3}

## The Sum-Difference Rule

If y = f(x) ± g(x)

## Differentiation Assignment Help By Online Tutoring and Guide Sessions at AssignmentHelp.Net

If y is the sum/difference of two or more functions of x: differentiate the 2 (or more) terms separately, then add/subtract.

(i) y = 2x^{2} + 3x then

dy/dx = 4 x + 3

(ii) y = 4x2 - x3 - 4x then

dy/dx = 8 x - 3 x ^{2} - 4

(iii) y = 5x + 4 then

dy/dx = 5

The Product Rule

If y = u.v where u and v are functions of x Then

i) y = (x+2)(ax^{2}+bx)

dy/dx = (x + 2)(2 ax + b)+(ax ^{2} + bx)

ii) y = (4x^{3}-3x+2)(2x^{2}+4x)

dy/dx = (4 x ^{3} - 3x +2)(2x^{2} + 4x)

dy/dx =(4x^{3} - 3 x + 2)(4 x + 4) + (2 x ^{2} + 4 x)(12 x^{2} - 3)

### The Quotient Rule

If y = u/v where u and v are functions of x

### The Chain Rule

If y is a function of v, and v is a function of x, then y is a function of x and

dy/dx = dy/dv . dv/dx

i) y = (ax^{2} + bx)

let v = (ax^{2} + bx), so y = v

dy/dx = 1/2 (ax ^{2} + bx )^{-1/2} . (2 ax + b)

ii) y = (4x^{3} + 3x 7 )^{4}

let v = (4x^{3} + 3x 7 ), so y = v^{4}

dy/dx = 4(4x^{3} + 3x - 7)^{3} . (12 x^{2} + 3)

The Inverse Function Rule

If x = f(y) then dy/dx = 1/ dx/dy

The derivative of the inverse of the function x = f(y), is the inverse of the derivative of the function

(i) x = 3y^{2} then

dy/dx = 6 y so dy/dx = 1/6y

(ii) y = 4x^{3} then

dy/dx = 12 x^{2} so dx/dy = 1/12 x^{2}

Rule 9: Differentiating Natural Logs

if y = loge x = ln x =>

dy/dx = 1/x

NOTE: the derivative of a natural log function does not depend on the co-efficient of x

Thus, if y = ln mx => dy/dx = 1/x

Proof

if y = ln mx m>0

**Rules of Logs** => y = ln m+ ln x

**Differentiating** (Sum-Difference rule)

dy/dx= 0 + 1/x = 1/x

Example:

Find the equation of the normal to the curve y = 3 x - x at the point where x = 4.

Solution:

The curve can be written as y = 3x^{1/2} - x

Therefore, dy/dx = 3/2 x^{-1/2} - 1

When x = 4, y = 34 - 4 = 2

and dy/dx = 3/2*4^{-1/2} -1 = 3/2*1/2*-1 = -1/4

So the gradient of the normal is m = -1/ -1/4 = 4 .

To find the equation of the tangent:

y = mx + c => y = 4x + c

Substitute x = 4, y = 2: 2 = 4(4) + c i.e. c = -14.

So equation is y = 4x 14.

### Important functions for differentiation:

Exponential and logarithmic functions:

d/dx e^{x} = e^{x}.

d/dx a^{x} = In(a)a^{x}.

d/dx In(x) = 1/x, x >0.

d/dx log_{a}(x) = 1/x In(a).

Trigonometric functions:

d/dx sin(x) = cos(x).

d/dx cos(x) = -sin(x).

d/dx tan(x) = sec^{2} = 1/cos^{2}(x) = 1 + tan^{2}(x).

Inverse trigonometric functions:

d/dx arcsin(x) = 1/1-x^{2}, -1 < x < 1.

d/dx arccos(x) = - 1/1-x^{2}, -1 < x < 1.

d/dx arctan(x) = 1/1 + x^{2}