



We can
find derivatives of things that aren't even functions. 


But to do
this, we need more special tricks. 





To make
the equations easier to read, 


we will use Y' to stand for the
derivative with these. 





Say you
have an equation like: 





X^{5}
+ 3XY^{2} = XY^{6} + X^{2}






It would
be pretty tough to solve this for Y. 


That is,
make it look like: 


Y
= something






Can we
ever find the equation for the derivative of the messy equation
above? 


WE CAN! 





The trick
is to thing of Y as being a function of X. 


We can't
EXPLICITLY work out what that function is, 


But we
can say it exists and use it IMPLICITLY as Y. 





So
anytime we have a Y, it is like the outside function 


with some
unknown function of X hiding inside. 


Look at
the second term of the messy equation above. 





3XY^{2}






Instead
of the way it's written, 


think of
it as: 


3X(some
messy function of X)^{2}






So we
have two things multiplied together, for that we need the product
rule. 


And one
of the things that are multiplied together 


is a function with an
outside and an inside part. 


For that,
we also need the power chain rule. 





Quickie
reminders: 


Product
Rule: 


If
F(x) = G(x) x H(x) then F'(x) = G(x) x H'(x) + G'(x) x
H(x)






Power
Chain Rule: 


If
F(x) = H(G(x))
then F'(x) = H'(G(x)) x G'(x)






So as we
go along working on the terms, 


the
derivative of X will be 1 AND the derivative of Y will be Y' 


If we
just worked this on the 3XY^{2} term, we would get: 





6XYY' + 3Y^{2}






OK, in
the second part (G'(X) +
H(X)), we had the 3, 


and the
derivative of the X term (that is, 1) 


and the Y
term (that is, Y^{ 2}) 





That
one's pretty easy to figure out. 





In the
first part, (G(X) +
H'(X)), we had the 3, 


and the X
term (that is, X) 


and the
derivative of the Y
term (that is, 2 YY') 


When you
put that all together, you get 6XYY', 


but lets look at the 2YY'
section again. 


Just how
did we get that? 





OK, we
start with Y^{ 2} and thing of it as (some messy X thing)^{
2}. 


When we
take the derivative, we get 2(some messy X thing) 


times the
derivative of the messy X thing. 


The name
for some messy X thing is Y. 


The name
for the derivative of some messy X thing is Y' 


Put that
all together, and we get 2YY'. 


Multiply
that by the 3 and the X and we get 6XYY'. 


And there
it is. 





So lets
take a run at the whole problem. 


Find the
derivative of: 


X^{5}
+ 3XY^{2} = XY^{6} + X^{2}






Taking
the derivative of each term: 





5X^{4}
+ 6XYY' + 3Y^{2}
= 6XY^{5}Y' + Y^{6} + 2X



What a
mess! 


Now what? 





Well,
we're looking for the derivative of Y, that is Y'. 


So put
all of the terms with a Y' on one side of the equation 


and
everything else on the other side. 





6XYY'  6XY^{5}Y'
= Y^{6}  3Y^{2}
 5X^{4} + 2X






Now just
get Y' by itself: 


Y'(6XY
 6XY^{5}) = Y^{6}
 3Y^{2}
 5X^{4} + 2X






Multiply
both sides by 6XY  6XY ^{5} 





Y'(6XY
 6XY^{5}) 

Y^{6}
 3Y^{2}
 5X^{4} + 2X 

= 

6XY
 6XY^{5} 

6XY
 6XY^{5} 






Y'
= 
Y^{6}
 3Y^{2}
 5X^{4} + 2X 

6XY
 6XY^{5} 






Nobody
said it was going to be pretty. 


And once
we get the first derivative, 


we can keep on going to the second
derivative and beyond. 


Don't
expect it to be any less messy as you go. 





Example: 





Find the
first and second derivatives of: 





X^{3}
+ 2Y^{3} = 43



Here we
go: 


3X^{2}
+ 6Y^{2}Y' = 0 


6Y^{2} Y' =  3X^{2} 

















OK,
that's the first derivative. 


Now let's
find the derivative of that. 


We'll
need the power chain rule to deal with the Y, 


and we'll
need the quotient rule to deal with the denominator. 





Quotient
Rule Mini Review: 





If
F(X) = 
G(X) 
Then
F'(X) = 
H(X)
x G'(X) 
H'(X)
x G(X) 


H(X) 
(H(X))^{2} 






So on
with the show: 


Y''
= 
2Y^{2}
(2X)  (4YY')(X^{2}) 

(2Y^{2})^{2} 






Multiply
out and simplify a bit 


Y''
= 

4XY^{2} + 4X^{2}YY' 

4Y^{4} 






Y''
= 
(
XY + X^{2}Y')4Y 

4
Y^{
4 }^{ 3} 






Y''
= 

XY + X^{2}Y' 

Y^{3} 






OK, so is
that second derivative any good to us with a Y' in it????? 





Probably
not. 


How do we
get rid of it? 





Easy. 


We know
that: 








So just
substitute that into the equation for Y' and clean things up a bit: 





Y''
= 
XY + X^{2}
( 

X^{2} 
) 

2Y^{2} 

Y^{3} 






Y''
= 
XY  ( 
X^{4} 
) 

2Y^{2} 

Y^{3} 






Y''
= 
 
2XY^{3} 


X^{4} 



2Y^{2} 
2Y^{2} 

Y^{3} 






Y''
= 

2XY^{3 }+ X^{4} 


2Y^{2} 

Y^{3} 






Y''
= 
2XY^{3 }+ X^{4}


2Y^{5} 






copyright 2005 Bruce Kirkpatrick 
