Introduction to differential calculus part 2

Geology 351 - Geomathematics

Tom Wilson, Department of Geology and Geography

[email protected]

Dept. Geology and GeographyWest Virginia University

Introduction to differential calculus – part 2

Last time


Basic differentiation rules:• Power rule• Sum rule• Chain rule• Trig functions• Product and quotient rules• Exponential rules• Logarithmic differentiation

More in-class example problems and brief assessment activity

At 2.5km, the slope is -0.05

at 3.5 km the slope is -0.0259

at 0.5, -0.191


Z

0 1 2 3 4 5

PH

I

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4Porosity depth relationship

=0.4e-z/1.5

Reviewed in last class


0.5~ 0.19

1.5~ 0.1

3~ 0.04

z

z

z

Slopes evaluated over z of

0.2km about a given depth

You’ll find the derivative z

ode

dz

Evaluation of the derivative at each of these points yields

0.51.5

1.51.5

1.51.5

0.40.19

1.5

0.4 0.098

1.5

0.4 0.036

1.5

zod

e edz

e

e

The individual slope calculations provide approximate estimates of the

slope at the midpoint. The derivative is exact and also provides an

analytical expression that is easily plotted.

What is the intercept for the derivative?


Z

0 1 2 3 4 5

PH

I

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4Porosity depth relationship

=0.4e-z/1.5

ozd

edz

The intercept would be o/ or, in this case 0.267km-1

A brief look at derivatives of trig functions.

Consider dsin()/d.

Start with the following -


sin( ) sin( )

sin( ) sin cos cos sin

sin( ) sin cos cos sin

cos( ) cos cos sin sin

cos( ) cos cos sin sin

identities

Take notes as we go through this and the derivative of the cosine in class.

We end up with


0

sincos lim

3 5 7

sin ...3! 5! 7!

When is small (such as in ), sin~

We can also see this graphically using arc-length relationships

Intro Continued – trig functions and the chain rule


)2sin( axy

let 2 so that sin(2 ) sin( )h ax ax h

then dx

dh

dh

dy

dx

dy.

Which reduces to aaxdx

dy2).2cos( or just

)2cos(2 axadx

dy

(the angle is another function 2ax)


In general if

))...))))((...(((( xqihgfy

then

dx

dq

di

dh

dh

dg

dg

df

df

dy

dx

dy........

The chain rule is a tool that can be used explicitly by defining h and then its

derivative. Alternatively, one often eliminates explicit definition of h and

implements the chain rule using, for example, an outside-to-inside

differentiation of terms.

Either way – one has to show all steps!


How do you handle derivatives of functions like

)()()( xgxfxy

?

or

)(

)()(

xg

xfxy

The products and quotients of other functions


fgy

Removing explicit reference to the independent variable x, we have

))(( dggdffdyy

Going back to first principles, we have

Evaluating this yields

dfdgfdggdffgdyy

Since dfdg is very small we let it equal zero; and since y=fg, the

above becomes -

Product rule applied to straight line formula

with its constants


y=ax+b

y

x

b is the intercept

The slope a=y/x is a constant

y'=xda/dx+adx/dx +db/dx = a

b, the intercept is a constant that just gets added to the ax and shifts it up or down. The slope does not change.

Product rule applied to y=ax+b 0 01The slope is

constant for all

values of intercept b


&dy df dg

dy gdf fdg g fdx dx dx

Which is a general statement of the rule used to evaluate the derivative of a product of functions.

The quotient rule is just a variant of the product rule, which is used to differentiate functions like

g

fy


2g

dxdg

fdx

dfg

g

f

dx

d

The quotient rule states that

The proof of this relationship can be tedious, but I think you can get it much easier using the power rule

Rewrite the quotient as a product and apply the product rule to y as shown below

1 fgg

fy


fhy

We could let h=g-1 and then rewrite y as

Its derivative using the product rule is just

dx

dhf

dx

dfh

dx

dy

dh = -g-2dg and substitution yields

2g

dxdg

f

g

dxdf

dx

dy


2g

dxdg

f

g

dxdf

g

g

dx

dy

Multiply the first term in the sum by g/g (i.e. 1) to get >

Which reduces to

2g

dxdg

fdx

dfg

dx

dy

the quotient rule

Rule for exponential differentiation discussed earlier


xxde

edx

( )

cxcx cxdAe d cx

Ae cAedx dx

This is an application of the rule for differentiating exponents and the chain rule

Basically indestructible in this form;

( )( )

h xh xde dh

edx dx

Rewrite the function

Multiply times derivative of the exponent

For a function like , this is not the case. Calculating the derivative becomes a little more complex.

cxAe

2

2, etc.

xxd e

edx


xxde

edx

( )cxcx cxdAe d cx

Ae cAedx dx

Basic rule for differentiating exponential functions

Sketch and discuss

Rewrite the exponential function and multiply it by the derivative of the exponent – a two-step process.

Second derivative?


2

2

cxd Ae

dx

cxcxdAe

cAedx

2

2

cx cxd Ae dcAe

dx dx and

Follow up on carrying the constants through


Use the product rule to differentiate a simple function like

2y ax

dy dg dff g

dx dx dx

….

Derivative of exponential function with base

other than e


The preceding rules for differentiation of exponential functions are

specific to base e.

An exponential function such as N=k10-bm must be converted into an

exponential function with base e.

This is easily done for any base. In the case of base 10 (or any base)

simply express 10 as e2.3026 where 2.3026 is the natural log of 10 (e.g.

ln10).

The above expression for N becomes 2.3026 2.3026 or bm

bmN k e ke

Thus

2.3026

2.30262.3026

bm

bmd kedN

b kedm dm

Differentiating exponential functions of

arbitrary base (cont.)


2.3026

2.30262.3026

bm

bmd kedN

b kedm dm

Since e2.3026 = 10 &

also since 2.3026=ln(10) we can rewrite the derivative

2.3026

2.3026 10

bm

bmd kedN

b kdm dm

This rule can be applied in general to any base and would

be an approach you could use for problem 13.

2.3026

ln(10 10)

bm

bmd kedN

b kdm dm

If you are cross-checking using WolframAlpha


note

Derivative of logarithmic functions


The derivative of ln(x) is 1/x. The application to more complex functions

follows the same general rule and incorporates use of other differentiation

rules as needed. As a general example take ln (f(x)).

ln( ( ))

1 ( )

( )

given y f x

dy df x

dx f x dx

This incorporates a chain rule application.

Depending on the nature of f(x), the chained component may require

use of other basic rules (e. g. product rule, sum rule, etc.).


If we finish these today

hand in otherwise bring in

for discussion next time

Next time we’ll put the rule to the test using Excel


0

czdc e

dz

In the lab exercise c = 1.

derivative

Before leaving answer the following

assessment questions


Please do these on your own. These are not group problems.

You will receive 7 points regardless, so work these independently. The

purpose of this assessment activity is to determine where additional focus

should be placed as we move forward.

You will receive feedback.

Take about 10 minutes


Next time we’ll spend some more time with

derivatives of exponential and log functions


xexi . )( 2

)sin(.3 )( 2 yii

)tan(.xx.cos(x) )( 2 xziii

24 17)ln(.3 )( Biv

Find the derivatives of

See Waltham problem 8.8

Before leaving


• Turn in the in-class worksheet• Hand in the assessment questions• That’s it for today

For next time - look over problems 8.13 and 8.14


•Bring questions to class (8.13, 8.14, …)•No due date set at present for these two problems.•Keep reviewing materials in chapter 8 to refresh your memory of basic differentiation rules and applications. There are several good examples in the text.

Introduction to differential calculus part 2

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Transcript of Introduction to differential calculus part 2