tom.h.wilson [email protected] dept. geology and geography west virginia university
TRANSCRIPT
X
0 2 4 6 8 10 12 14
y = x3
0
500
1000
1500
2000
X
0 5 10 15 20
y = x2
0
50
100
150
Slope
X2 X1 del x del y slope7 5 2 24 128 4 4 48 12
10 2 8 96 12
X2 X1 del x del y slope7 5 2 218 1098 4 4 448 112
10 2 8 992 124
Estimating the rate of change of functions with variable slope
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
432234 )()(4)(64 dxdxxdxxdxxxdyy
Remember the idea of the dy and dx is that they represent differential changes that 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 very 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 -
34xdx
dy
Divide both sides of this equation by dx to get
dxxdy 34
This is just another illustration of what you already know as the power rule,
1 nnaxdx
dyis
Just as a footnote, 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( xadx
dy
Examining the effects of differential increments in y and x we get the
following
Don’t let negative exponents fool you. If n is -1, for example, we still have
1 nnaxdx
dy
2 axdx
dy
or just
)()()( xgxfxy Given the function -
what is dx
dy?
dx
dg
dx
df
dx
dy
We just differentiate f and g individually and take their sum, so that
Take the simple example )()( 42 baxcxy
- what is
dx
dy?
baxcxy 42We can rewrite
Then just think of the derivative operator as being a distributive operator that acts on each term in the sum.
Where
then -
On the first term apply the power rule
What happens to
baxcxy 42
2 4( )dy d
x c ax bdx dx
2 4dy dx dc dax db
dx dx dx dx dx an
d2dx
dxdc
dx?
Successive differentiations yield
Thus -
342 axxdx
dy
2 4
32 0 4 0
dy dx dc dax db
dx dx dx dx dxdy
x axdx
)()( 42 baxcxdx
d
dx
dy
Differences are treated just like sums
so that
is just
342 axxdx
dy
Differentiating functions of functions -
Given a function
22 )1( xy we consider
)()1( 2 xhx write 2hy compute
hhdh
d
dh
dy22
Then compute
xxdx
d
dx
dh212 an
d
take the product of the two, yielding dx
dh
dh
dy
dx
dy.
xxdx
dh
dh
dy
dx
dy2).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
xxdx
dy2.)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 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........
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 -
fdggdfdy
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
2gdx
dgfdxdfg
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 fg
g
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
2gdx
dgf
gdx
df
dx
dy
2gdx
dgf
gdx
df
g
g
dx
dy
Multiply the first term in the sum by g/g (i.e. 1) to get >
Which reduces to
2gdx
dgfdxdfg
dx
dy
the quotient rule
Z (km)0 1 2 3 4 5
0.0
0.1
0.2
0.3
0.4
0.5Porosity-Depth Relationship
Slope
cxAeFunctions of the type
czoe Recall our earlier discussions
of the porosity depth relationship
Z (km)0 1 2 3 4 5
0.0
0.1
0.2
0.3
0.4
0.5Porosity-Depth Relationship
Slope
czoe
?z
Refer to comments on the computer lab exercise.
Derivative concepts
Z (km)0 1 2 3 4 5
0.0
0.1
0.2
0.3
0.4
0.5Porosity-Depth Relationship
Slope
czoe
?z
Between 1 and 2 kilometers the gradient is -0.12 km-1
Z (km)
0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2
0.16
0.18
0.20
0.22
0.24
0.26
0.28
0.30
0.32
0.34Porosity-Depth Relationship
Gradient1 to 2 km
Gradient1.0 to 1.1 km
As we converge toward 1km, /z decreases to -0.14 km-1 between 1 and 1.1 km depths.
Z (km)
0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2
0.16
0.18
0.20
0.22
0.24
0.26
0.28
0.30
0.32
0.34Porosity-Depth Relationship
Gradient1 to 2 km
Gradient1.0 to 1.1 km
What is the gradient at 1km?
What is ?d
dz
xxdee
dx
( )cxcx cxdAe d cx
Ae cAedx dx
This is an application of the rule for differentiating exponents and the chain
rule
Next time we’ll continue with exponentials and logs, but also have a look at 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
Differentiating exponential and log functions
The log and its derivative
-10
-5
0
5
10
0 10 20 30 40 50
X
Yln(x)
1/x
Z (km)0 1 2 3 4 5
0.0
0.1
0.2
0.3
0.4
0.5Porosity-Depth Relationship
Slope
axey ( ) axdy d axe
dx dx axae
Returning to those exponential and natural log cases – recall we implement the chain rule when differentiating
h in this case would be ax and, from the chain rule,
dx
dh
dh
dy
dx
dy. become
s dx
dh
dh
de
dx
dy h
. or
dx
dhe
dx
dy h. and finally
axaedx
dy
since
axh and
adx
dh
For functions like 2axey
we follow the same procedure.
Let 2axh and then
From the chain rule we have dx
dh
dh
dy
dx
dy.
axdx
dh2
hh eedh
d
dh
dy
22. axaxe
dx
dh
dh
dy
dx
dyhence
Thus for that porosity depth relationship we were working with
- /
0)( zez
?
)( /0
dz
ed
dz
zd z
/0 ze
•The derivative of logarithmic functions
Given >
)ln(xy
xdx
dy 1
We’ll talk more about these special cases after we talk about the chain
rule.
For logarithmic functions like )ln( 2xy
We combine two rules, the special rule for natural logs and the chain rule.
Let 2xhdx
dh
dh
dy
dx
dy.Chain
rule
Log
rule xdx
xd 1)(ln
then
2
1
xdh
dy an
dx
dx
dh2
soxx
x
dx
xd 22)ln(2
2
•The derivative of an exponential
function
xey
xedx
dy
Given >
In general for axey xedx
axd
dx
dy )( axae
xay If express a as en so that nxxn eey
then nxnx needx
d
dx
dy
)ln()ln( aen n Note
nxnx needx
d
dx
dy
Since nxx ea and
)ln(an
xaadx
dy . )ln(in general
a can be thought of as a general base. It could be 10 or 2, etc.