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We can
find derivatives of things that aren't even functions. |
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But to do
this, we need more special tricks. |
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To make
the equations easier to read, |
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we will use Y' to stand for the
derivative with these. |
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Say you
have an equation like: |
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X5
+ 3XY2 = XY6 + X2
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It would
be pretty tough to solve this for Y. |
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That is,
make it look like: |
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Y
= something
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Can we
ever find the equation for the derivative of the messy equation
above? |
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WE CAN! |
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The trick
is to thing of Y as being a function of X. |
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We can't
EXPLICITLY work out what that function is, |
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But we
can say it exists and use it IMPLICITLY as Y. |
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So
anytime we have a Y, it is like the outside function |
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with some
unknown function of X hiding inside. |
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Look at
the second term of the messy equation above. |
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3XY2
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Instead
of the way it's written, |
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think of
it as: |
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3X(some
messy function of X)2
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So we
have two things multiplied together, for that we need the product
rule. |
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And one
of the things that are multiplied together |
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is a function with an
outside and an inside part. |
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For that,
we also need the power chain rule. |
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Quickie
reminders: |
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Product
Rule: |
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If
F(x) = G(x) x H(x) then F'(x) = G(x) x H'(x) + G'(x) x
H(x)
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Power
Chain Rule: |
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If
F(x) = H(G(x))
then F'(x) = H'(G(x)) x G'(x)
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So as we
go along working on the terms, |
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the
derivative of X will be 1 AND the derivative of Y will be Y' |
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If we
just worked this on the 3XY2 term, we would get: |
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6XYY' + 3Y2
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OK, in
the second part (G'(X) +
H(X)), we had the 3, |
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and the
derivative of the X term (that is, 1) |
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and the Y
term (that is, Y 2) |
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That
one's pretty easy to figure out. |
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In the
first part, (G(X) +
H'(X)), we had the 3, |
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and the X
term (that is, X) |
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and the
derivative of the Y
term (that is, 2 YY') |
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When you
put that all together, you get 6XYY', |
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but lets look at the 2YY'
section again. |
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Just how
did we get that? |
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OK, we
start with Y 2 and thing of it as (some messy X thing)
2. |
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When we
take the derivative, we get 2(some messy X thing) |
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times the
derivative of the messy X thing. |
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The name
for some messy X thing is Y. |
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The name
for the derivative of some messy X thing is Y' |
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Put that
all together, and we get 2YY'. |
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Multiply
that by the 3 and the X and we get 6XYY'. |
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And there
it is. |
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So lets
take a run at the whole problem. |
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Find the
derivative of: |
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X5
+ 3XY2 = XY6 + X2
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Taking
the derivative of each term: |
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5X4
+ 6XYY' + 3Y2
= 6XY5Y' + Y6 + 2X
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What a
mess! |
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Now what? |
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Well,
we're looking for the derivative of Y, that is Y'. |
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So put
all of the terms with a Y' on one side of the equation |
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and
everything else on the other side. |
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6XYY' - 6XY5Y'
= Y6 - 3Y2
- 5X4 + 2X
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Now just
get Y' by itself: |
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Y'(6XY
- 6XY5) = Y6
- 3Y2
- 5X4 + 2X
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Multiply
both sides by 6XY - 6XY 5 |
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Y'(6XY
- 6XY5) |
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Y6
- 3Y2
- 5X4 + 2X |
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6XY
- 6XY5 |
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6XY
- 6XY5 |
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| Y'
= |
Y6
- 3Y2
- 5X4 + 2X |
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| 6XY
- 6XY5 |
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Nobody
said it was going to be pretty. |
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And once
we get the first derivative, |
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we can keep on going to the second
derivative and beyond. |
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Don't
expect it to be any less messy as you go. |
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Example: |
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Find the
first and second derivatives of: |
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X3
+ 2Y3 = 43
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Here we
go: |
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3X2
+ 6Y2Y' = 0 |
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6Y2 Y' = - 3X2 |
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OK,
that's the first derivative. |
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Now let's
find the derivative of that. |
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We'll
need the power chain rule to deal with the Y, |
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and we'll
need the quotient rule to deal with the denominator. |
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Quotient
Rule Mini Review: |
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| If
F(X) = |
G(X) |
Then
F'(X) = |
H(X)
x G'(X) -
H'(X)
x G(X) |
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| H(X) |
(H(X))2 |
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So on
with the show: |
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Y''
= |
2Y2
(-2X) - (4YY')(-X2) |
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| (2Y2)2 |
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Multiply
out and simplify a bit |
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Y''
= |
-
4XY2 + 4X2YY' |
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| 4Y4 |
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Y''
= |
(-
XY + X2Y')4Y |
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4
Y
4 3 |
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OK, so is
that second derivative any good to us with a Y' in it????? |
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Probably
not. |
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How do we
get rid of it? |
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Easy. |
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We know
that: |
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So just
substitute that into the equation for Y' and clean things up a bit: |
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| Y''
= |
-XY + X2
( |
-
X2 |
) |
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| 2Y2 |
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| Y3 |
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| Y''
= |
-XY - ( |
X4 |
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| 2Y2 |
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| Y3 |
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| Y''
= |
- |
2XY3 |
-
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X4 |
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2Y2 |
2Y2 |
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| Y3 |
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copyright 2005 Bruce Kirkpatrick |
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