tom.h.wilsonwilson@geo.wvu.e

duDept. Geology and

GeographyWest Virginia University

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0.5Porosity-Depth Relationship

Slope

cxAeFunctions of the type

czoe Recall our earlier discussions

of the porosity depth relationship

xxde e

dx

( )cxcx cxdAe d cxAe cAe

dx dx

But it worked out a little differently for

xxde e

dx

Don’t get the derivatives of these exponential functions confused with the power rule

The book works through the differentiation of y = x2, so let’s try y =x4.

4)( dxxdyy

multiplying that out -- you get ... 432234 )()(4)(64 dxdxxdxxdxxxdyy

432234 )()(4)(64 dxdxxdxxdxxxdyy

Remember this idea of dy and dx is that the differential changes are infinitesimal - very small. So if dx is 0.0001 (that’s 1x10-4) then (dx)2 = 0.00000001 (or 1x10-8) (dx)3 = 1x10-12 and (dx)4 = 1x10-16.

So even though dx is infinitesimally small, (dx)2 is orders of magnitude smaller

432234 )()(4)(64 dxdxxdxxdxxxdyy

so that we can just ignore all those terms with (dx)n where n is greater than 1.

dxxxdyy 34 4

Our equation gets simple fast

Also, since y =x4, we have dxxydyy 34

dxxdy 34

and then -

34xdxdy

Divide both sides of this equation by dx to get

dxxdy 34

This illustrates the general form of the

power rule –

1 nnaxdxdyis

Again, as a reminder, remember that the constant factors in an expression carry through the differentiation.This is obvious when we consider the derivative -

baxy 2

which - in general for

naxy

bdxxadyy 2)(

bdxxdxxadyy )2( 22

axdxbaxdyy 2)( 2 axdxydyy 2

)2( xadxdy

Examining the effects of differential increments in y and x we get the

following

Don’t let negative exponents fool you either. If n is -1, for example, we still

have 1 nnax

dxdy

2 axdxdy

And the result is

dxaxd

dxdy )( 1

In this case

)()()( xgxfxy Given the function -

what is dxdy ?

dxdg

dxdf

dxdy

We just differentiate f and g individually and take their sum, so that

Take the simple example )()( 42 baxcxy

- what is dx

dy ?

What are the individual derivatives of )( 2 cx )( 4 bax and ?

)( 2 cxf let

then - dxcxd

dxdf )( 2

We just apply the power rule and obtain

xdxdf 2

We know from the forgoing note that the c disappears.

We use the power rule again to evaluate the second term, letting

g = (ax4+b)34axdxdg

Thus - 342 axx

dxdy

)()( 42 baxcxdxd

dxdy

Differences are treated just like sums

so that

is just 342 axx

dxdy

Recall how to handle derivatives of products

)()()( xgxfxy

?or

)()()(xgxfxy

fgy

Removing explicit reference to the independent variable x, we have

))(( dggdffdyy Going back to first principles, we have

Evaluating this yields dfdgfdggdffgdyy

Since df x dg is very small and since y=fg, the above becomes -

fdggdfdy

Which is a general statement of the rule used to evaluate the derivative of a product of functionsThe quotient rule is just a variant of the product rule, which is used to differentiate functions like

gfy

2gdx

dgfdxdfg

gf

dxd

The quotient rule states that

And in most texts the proof of this relationship is a rather tedious one.The quotient rule is easily

demonstrated however, by rewriting the quotient as a product and applying the product rule. Consider

1 fggfy

fhy

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

Its derivative using the product rule is just

dxdhf

dxdfh

dxdy

and substitution yields

2gdx

dgf

gdx

df

dxdy

It’s actually necessary to incorporate the chain rule into this

since dh dh dgdx dg dx

2dh gdg

2gdx

dgf

gdx

df

gg

dxdy

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

Which reduces to

2gdx

dgfdxdfg

dxdy

i.e. the quotient rule

•The derivative of an exponential

function

xey

xedxdy

Given >

In general for axey xedxaxd

dxdy )(

axae

xay If express a as en so that nxxn eey

then nxnx needxd

dxdy

)ln()ln( aen n Note

nxnx needxd

dxdy

Since nxx ea and

)ln(an

xaadxdy . )ln(in general

a can be thought of as a general base. It could be 10 or 2, etc.

•The derivative of logarithmic functions

Given >

)ln(xy

xdxdy 1

We’ll talk more about these special cases after we talk about the chain

rule.

Differentiating functions of functions -

Given a function

22 )1( xy we consider )()1( 2 xhx write 2hy comput

e hh

dhd

dhdy 22

Then compute

xxdxd

dxdh 212 an

dtake the product of the two, yielding dx

dhdhdy

dxdy .

xxdxdh

dhdy

dxdy 2).1(2. 2

)1(4 2 xx

22 )1( xy

We can also think of the application of the chain rule especially when powers are involved as working form the outside to inside of a function

22 )1( xyWhere

xxdxdy 2.)1(2 12

Derivative of the quantity squared viewed from the outside.

Again use power rule to differentiate the inside term(s)

Using a trig function such as )2sin( axy

let axh 2

then dxdh

dhdy

dxdy .

Which reduces to aaxdxdy 2).2cos( or just

)2cos(2 axadxdy

In general if))...))))((...(((( xqihgfy

then

dxdq

didh

dhdg

dgdf

dfdy

dxdy ........

axey ( ) axdy d ax edx dx

axae

Returning to those exponential and natural log cases - we already implemented the chain rule when differentiating

h in this case would be ax and, from the chain rule,

dxdh

dhdy

dxdy . become

s dxdh

dhde

dxdy h

. or

dxdhe

dxdy h. and

finally axae

dxdy

since

axh and

adxdh

For functions like 2axey

we follow the same procedure.

Let 2axh and then

From the chain rule we have dxdh

dhdy

dxdy .

axdxdh 2

hh eedhd

dhdy

22. axaxe

dxdh

dhdy

dxdy

hence

Thus for that porosity depth relationship we were working with

- /

0)( zez

?)( /0

dzed

dzzd z

/0 ze

For logarithmic functions like )ln( 2xy

We combine two rules, the special rule for natural logs and the chain rule.

Let 2xhdxdh

dhdy

dxdy .Chain

rule

Log

rule xdxxd 1)(ln

then

21xdh

dy an

dx

dxdh 2

soxx

xdxxd 22)ln(

2

2

For next Tuesday answer question 8.8 in Waltham (see page 148).

xexi . )( 2

)sin(.3 )( 2 yii

)tan(.xx.cos(x) )( 2 xziii 24 17)ln(.3 )( Biv

Find the derivatives of