# 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 = axn, a & n are constants

dy/dx = n . a . xn-1

i) y = 4x =>

dy/dx = 4 x 0 = 4

ii) y = 4x2 =>

dy/dx = 8 x

iii) y = 4x3 =>

dy/dx = 12 x2

iv) y = 4x-2 =>

dy/dx = - 8 x - 3

## The Sum-Difference Rule

If y = f(x) ± g(x) 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 = 2x2 + 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)(ax2+bx)

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

ii) y = (4x3-3x+2)(2x2+4x)

dy/dx = (4 x 3 - 3x +2)(2x2 + 4x)

dy/dx =(4x3 - 3 x + 2)(4 x + 4) + (2 x 2 + 4 x)(12 x2 - 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 = (ax2 + bx)½

let v = (ax2 + bx), so y = v½

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

ii) y = (4x3 + 3x – 7 )4

let v = (4x3 + 3x – 7 ), so y = v4

dy/dx = 4(4x3 + 3x - 7)3 . (12 x2 + 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 = 3y2 then

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

(ii) y = 4x3 then

dy/dx = 12 x2 so dx/dy = 1/12 x2

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 = 3x1/2 - x

Therefore, dy/dx = 3/2 x-1/2 - 1

When x = 4, y = 3√4 - 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 ex = ex.

d/dx ax = In(a)ax.

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

d/dx loga(x) = 1/x In(a).

Trigonometric functions:

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

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

d/dx tan(x) = sec2 = 1/cos2(x) = 1 + tan2(x).

Inverse trigonometric functions:

d/dx arcsin(x) = 1/√1-x2, -1 < x < 1.

d/dx arccos(x) = - 1/√1-x2, -1 < x < 1.

d/dx arctan(x) = 1/1 + x2