Differentiation and Integration

C3 – CALCULUS (Differentiation and Integration)

Differentiation

If y =

[Remember, =gradient of tangent to curve y = f(x) ]

Revision examples: Differentiate the following functions;

i) ii) iii) iv) y= v) vi)

vii) vii)

Solution

i)

ii)

iii)

iv)

v)

vi)

vii)

viii)

/tt/file_convert/546a9ca8b4af9ff1268b4818/document.docXaverian Page 1 4/8/2023


Another Revision Example:

Given that , find and hence find the gradient of the curve at the

point (9,27)

Solution

When , so gradient at the point is

The Chain Rule

Used to differentiate a composite function (i.e. one function inside another

function) e.g. .

We start by letting t = the inner function: in the above example t =

The chain rule says:

Example: Differentiate

Solution

Let t = so



Example: Use the chain rule to find when

Solution:

NB: To apply the chain rule quickly remember:

Exercise: C3/4 textbook, Ex 1, P.69, Q. 2(a,c), 3(a,c,e), 4, 8



Connected Rates of Change

The rate of change of x is , where t is time. If we know a formula for

in terms of t, we find the rate of change of by differentiation.

NB: If is increasing at 6cm s-1 then = + 6,

if is decreasing at 6cm s-1 then

We can sometimes use the Chain Rule to find the rate of change of a variable if we know the rate of change of a related variable.

Example: Find the rate of increase of the radius of a sphere whose volume is increasing at a rate of 100 cm3 s-1 at the instant when the radius is 3cm.

Solution:

We know = 100 and we want to find .when r=3

Using the Chain Rule,

We therefore need to find a relationship between V and r so we can find

For a sphere, Volume, V=

So,

Therefore,

(when the radius is 3cm)

C3/4 textbook: Ex 3, p159, Q. 6, 7, 9, 10,13, 14



The product rule

If y = uv (where u and v are functions of x), then

Example 1

Given that , find using the product rule:

Solution

Example 2: Find if

Solution: (we need to use both the chain and product rules for this)

(using chain rule)_______________________________________________

(using product rule)



Example 3: Find the x-coordinates of the stationary points of the curve

Solution:

Then

Exercise

Differentiate the following using the product rule:

i)

ii)

iii) , and find the coordinates of any stationary points

iv)

v)

vi) …………do you get the same answer as when we multiplied it out first ?

The quotient rule/tt/file_convert/546a9ca8b4af9ff1268b4818/document.docXaverian Page 6 4/8/2023


If y = (where u and v are functions of x), then

Example 1:

Given that , (i) find and hence (ii) find the equation of the

tangent to the curve at the point (1, )

Solution:

(ii) When

Using ,

is the equation of the tangent.



Example 2: Given that find

Solution: (we need to use both the chain and quotient rules for this)

Exercise

1. Differentiate the following using the Quotient Rule:

i)

ii)

iii)

2. Find the equation of the tangent to the curve at the point where

=



Integration

NB

Example 1:

Solution

Example 2: Evaluate

Solution:



Example 3: Find the area represented by;

Solution;

Example 4:

i) Find the general solution of the differential equation.ii) Find the equation of the curve with this gradient function which

passes through (1,5)



Solution: (i)

(ii) Curve passes through (1,5) sub into

Example 5:

Solution: There is no quotient rule for integration so we must divide first before integrating.



Integration involving a Linear Substitution

If we have a composite function where the inner function is linear, we can always integrate by substitution.

Steps: 1) Use the substitution ‘let u = the linear function’.

2) Find and express in terms of

3) Substitute u for in the function to be integrated.4) Substitute for its equivalent in terms of .5) Substitute any ‘stray’ terms if necessary.6) Carry out the integration .7) Replace u with equivalent.

Example 1: Evaluate

Solution:

Example 2: Evaluate



Solution

Harder Integration by Substitution (with ‘stray’ x terms)

Example 1

Find

Solution Let u =

Then replace each with its equivalent in terms of u.

i.e …….multiply out before integrating….



………..replace the x…………….

Example 2

Find

Solution:

Let u =

Then replace each and multiply out before integrating

…….replace the x…………

Example 3

Evaluate

Solution:

Let u

Then replace each :

….multiply out before integrating….



……replace the x………….

Volumes of Revolution : When the area between a curve and the x-axis is rotated through one revolution (360o) about the x-axis a volume of revolution is formed.

If the curve y=f(x) from x=a to x=b is rotated about the x-axis through 360o , the volume formed is given by

V=

[About the y-axis, from y=p to y=q, V = ]

Example 1: Find the volume of revolution formed when the area enclosed

by the curve y= , the x-axis and the lines x=1 and x=3 is rotated 360o

about the x-axis.Solution:



Example 2: Find the volume of revolution formed when the curve 2y = x -1 is rotated through one revolution about the y axis from y=1 to y=4

Solution:

Since the rotation is about the y-axis, we need to get x in terms of y.

Rearranging gives x = 2y + 1


Documents

Differentiation and Integration