Calculus Implicit Differentiation
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When You Just Can't Shake the Y
Implicit Differentiation

 
 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:
 

 X5 + 3XY2 = XY6 + X2

 
 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.
 

 3XY2

 
 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 3XY2 term, we would get:
 

6XYY' + 3Y2

 
 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:

 X5 + 3XY2 = XY6 + X2

 
 Taking the derivative of each term:
 

 5X4 + 6XYY' + 3Y2 = 6XY5Y' + Y6 + 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' - 6XY5Y' =  Y6 - 3Y2 - 5X4 + 2X

 
 Now just get Y' by itself:

 Y'(6XY - 6XY5) =  Y6 - 3Y2 - 5X4 + 2X

 
 Multiply both sides by 6XY - 6XY 5
 
Y'(6XY - 6XY5 Y6 - 3Y2 - 5X4 + 2X

=
6XY - 6XY5 6XY - 6XY5
 
Y' =  Y6 - 3Y2 - 5X4 + 2X

6XY - 6XY5
 
 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:

 X3 +  2Y3 = 43

 Here we go:

3X2 +  6Y2Y' = 0

  6Y2 Y' = - 3X2
 
Y' = - 3X2

6 Y2
 
Y' = - X2

2Y2
 
 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'' = 

2Y2 (-2X) - (4YY')(-X2)

(2Y2)2
 
 Multiply out and simplify a bit

Y'' = 

- 4XY2 + 4X2YY'

4Y4
 

Y'' = 

(- XY + X2Y')4Y

 4 Y 3
 

Y'' = 

- XY + X2Y'

Y3

 

 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:
Y' = - X2

2Y2
 
 So just substitute that into the equation for Y' and clean things up a bit:
 
Y'' =

-XY + X2 (

- X2

)


2Y2

Y3
 
Y'' =

-XY - (

X4

)


2Y2

Y3

 

Y'' =

-

2XY3 - X4



2Y2 2Y2

Y3
 
Y'' = 2XY3 + X4


2Y2

Y3
 
Y'' = 2XY3 + X4

2Y5
 

   copyright 2005 Bruce Kirkpatrick

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