MODULE 3

DIFFERENTIATION

Derivative of a Function:

The derivative of a function f with respect to the variable x is defined as

provided the limit exists.

A function that is differentiable at every point of its domain is a differentiable function.

Example 1:Find the derivative of ( or differentiate)

Derivative at a point:

The derivative of the function f at the point x = a is the limit

if it exists.

Example 2: Use the above definition to find the derivative of

at x = a.

Notation:

Let y = f(x)

The derivative of f with respect x is written in several different ways:

Note: The derivative of f at x =a, if it exists, is the slope of the tangent at the point

(a, f(a)).

Exercise:Verify that f(x) = |x|, the absolute value function, does not have a derivative at

x = 0.

does not exist.

Differentiability implies Continuity:

If f has a derivative at x =a, then f is continuous at x =a.

Now

Note: The derivative of a constant function f(x) = c is 0.

Derivative of the Power Function:

Thus

Rules for Finding Derivatives:

·If f(x) = c, a constant function, then its derivative is 0

Product Rule:

Quotient Rule:

Find the first and second derivatives:

Multiply the right side:

Using the product rule:

Using Quotient Rule:

5. Find the equation of the tangent and normal to following the curve at (2,3):

Slope of the tangent = 3(2)^2 – 3 = 9

Slope of the normal = -1/9

Equation of the tangent at (2,3) is: y – 3 = 9(x – 2)

Equation of the normal at (2,3) is : y – 3 = -1/9(x – 2)

Derivatives of Trigonometric Functions:

Find the derivative of y = sin(x)

Its limit as h approaches 0 is cos(x)

Find the derivatives:

The Chain Rule and Parametric Equations:

If u is a differentiable function at x and f is differentiable at u(x), then the composite function f(u(x)) is differentiable at x and

Examples:

Implicit Differentiation:

Use implicit differentiation to find dy/dx:

Take x =1 and y = 0

6.Find the equation of the tangent at (2,1) to the curve given by

Substitute x = 2 and y = 1.

The equation of the tangent is y – 1 =3(x – 2).

7.Particle motion:

The position of a particle moving on a number line is given by:

where t is in seconds and s is in feet.

When is the particle stand still? moving to the left? moving to the right?

Graph its velocity and speed.

velocity = v = ds/dt = 6t^2 - 42t + 60

= 6(t –2)(t – 5)
The particle is stand still when v = 0
i.e. t = 2,5
It is moving left when v is negative and moving right when v is positive.
c.When is its acceleration positive, negative or zero?
It acceleration a(t) = dv/dt =12 t -42
(see the graph on the left for answer)

The distance
s = area below the velocity curve

s(2) = area below the velocity curve
            over [0,2].

To calculate s(6), all the areas over [0,6] have to be added using proper signs.

velocity

v = area below the acceleration curve.

For example velocity v when t = 2

v(2) = initial velocity + area below the
           acceleration curve over [0,2]

       = 60 – 2((42+18)/2) = 0