stroud worked examples and exercises are in the text programme f11 differentiation

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Worked examples and exercises are in the text

PROGRAMME F11

DIFFERENTIATION

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The gradient of a straight-line graph

The gradient of a curve at a given point

Algebraic determination of the gradient of a curve

Derivatives of powers of x

Differentiation of polynomials

Derivatives – an alternative notation

Differentiation of products of functions

Differentiation of a quotient of two functions

Functions of a function

Second derivatives

Newton-Raphson iterative method [optional]

Programme F11: Differentiation

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The gradient of the sloping line straight line in the figure is defined as:

the vertical distance the line rises and falls between the two points P and Qthe horizontal distance between P and Q

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The gradient of the sloping straight line in the figure is given as:

and its value is denoted by the symbol dy

mdx

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Second derivatives

Newton-Raphson iterative method


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The gradient of a curve between two points will depend on the points chosen:

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The gradient of a curve at a point P is defined to be the gradient of the tangent at that point:

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(Second derivatives –MOVED to a later set of slides)



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The gradient of the chord PQ is and the gradient of the tangent at P is

y

x

dy

dx

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As Q moves to P so the chord rotates. When Q reaches P the chord is coincident with the tangent.

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Two straight lines

Two curves


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Two straight lines


(a) (constant)y c

0 therefore 0dy

dydx

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Two straight lines

(b) y ax

. therefore dy

dy a dx adx

( ) y dy a x dx


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Two curves


(a)

so

2 y x

therefore 2dy

xdx

2( ) y y x x

22 . therefore 2

yy x x x x x

x

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Two curves


(b)

so

3 y x

2therefore 3dy

xdx

3( ) y y x x

2 32

22

3 . 3 .

therefore 3 3 .

y x x x x x

yx x x x

x

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A clear pattern is emerging:

1If then n ndyy x nx

dx

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At Q:

So

As

Therefore

called the derivative of y with respect to x.

22 5y y x x

222 4 . 2 5x x x x

24 . 2 and 4 2.

yy x x x x x

x

0 so the gradient of the tangent at y dy

x Px dx

4dy

xdx

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To differentiate a polynomial, we differentiate each term in turn:

4 3 2

3 2

3 2

If 5 4 7 2

then 4 5 3 4 2 7 1 0

Therefore 4 15 8 7

y x x x x

dyx x x

dx

dyx x x

dx

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The double statement:

can be written as:

4 3 2

3 2

4 3 2 3 2

If 5 4 7 2

then 4 5 3 4 2 7 1 0

5 4 7 2 4 15 8 7

y x x x x

dyx x x

dx

dx x x x x x x

dx

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Towards derivatives of trigonometric functions (JAB)

Limiting value of

is 1

I showed this in an earlier lecture by a rough argument.

Following slide includes most of a rigorous argument.


sin0 as

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Area of triangle POA is:

Area of sector POA is:

Area of triangle POT is:

Therefore:

That is ((using fact that the cosine tends to 1 -- JAB)):

212 sinr

212 r

212 tanr

2 2 21 1 12 2 2

sinsin tan so 1 cosr r r

0

sin1Lim

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Derivatives of trigonometric functions and …


The table of standard derivatives can be extended to include trigonometric and the exponential functions:

sincos

cossin

xx

d xx

dx

d xx

dx

dee

dx

((JAB:)) The trig cases use the identities for finding sine and cosine of the sum of two angles, and an approximation I gave earlier for the cosine of a small angle. (Shown in class).

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Given the product of functions of x:

then:

This is called the product rule.

y uv

dy dv duu v

dx dx dx

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Given the quotient of functions of x:

then:

This is called the quotient rule.

uy

v

2

du dvv udy dx dx

dx v

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Differentiation of a function of a function

To differentiate a function of a function we employ the chain rule.

If y is a function of u which is itself a function of x so that:

Then:

This is called the chain rule.


( ) ( [ ])y x y u x

dy dy du

dx du dx

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Differentiation of a function of a function


Many functions of a function can be differentiated at sight by a slight modification to the list of standard derivatives:

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Newton-Raphson iterative method [optional]


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Newton-Raphson iterative method [OPTIONAL]

Tabular display of results


Given that x0 is an approximate solution to the equation f(x) = 0 then a better solution is given as x1, where:

This gives rise to a series of improving solutions by iteration using:

A tabular display of improving solutions can be produced in a spreadsheet.

01 0

0

( )

( )

f xx x

f x

1

( )

( )n

n nn

f xx x

f x

stroud worked examples and exercises are in the text programme f11 differentiation

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