Implicit vs. Explicit Differentiation

Date: 01/04/2001 at 20:18:35
From: Meghan Fraser
Subject: Implicit vs. Explicit Differentiation

What is the exact definition of 'Implicit Differentiation'? I have a 
very hard time with math terms (please use layman's terms). What is 
'Explicit Differentiation'? I don't even know what the words implicit 
and explicit mean! What is the difference between them? Which do you 
use when?

Thank you for the help.

Date: 01/05/2001 at 17:26:09
From: Doctor Fenton
Subject: Re: Implicit vs. Explicit Differentiation

Dear Meghan,

Thanks for writing to Dr. Math. This is a good question, and it 
depends on remembering what we mean by a function. Recall that a 
function is a rule that assigns exactly one y-value to a given x. 
Another way of stating this is that a vertical line always intersects 
the graph of a function in only one point.

A function defined in the form

     y = f(x)

is said to be "explicitly" defined. That means that

     y = x^2   or  y = sin(x)

is explicitly defined.

Sometimes we have a graph given in a form that isn't, or perhaps 
can't be, solved for y. For example:

     x^2 + y^2 = 1

or

     y*e^y = x

In the first case, you probably recognize the equation as the equation 
of a circle centered at the origin of radius 1. The whole graph is not 
a function because vertical lines between x = -1 and x = 1 hit the 
circle twice. In this example, we actually can solve for y:

     y^2 = 1 - x^2

so either

     y = + sqrt(1 - x^2)

or

     y = - sqrt(1 - x^2)

Both of these formulas define explicit functions: y = + sqrt(1 - x^2) 
is the upper half of the circle; and the other formula defines the 
lower half of the circle. Each semicircle IS the graph of a function, 
because vertical lines now hit the graph only once.

We say that a function graph that is part of a larger graph (which may 
not be a function graph in its entirety) is "implicitly" defined by 
the larger graph or its formula.

The second example gives a graph that IS a function graph, but there 
is just no way to rewrite it in the form y = f(x), so we say in this 
case, too, that y is "implicitly" defined by the equation (y*e^y = x 
in this case).

For implicit differentiation, we have a formula

     F(x,y) = 0

which we either can't or don't bother to solve for y. For example, the 
circle equation would be:

      x^2 + y^2 - 1 = 0

and the other equation would be:

     y*e^y - x = 0

We can still compute derivatives, by using the Chain Rule whenever y 
occurs in the formula. In the circle example, the left side is a sum 
of three functions of x: x^2, y^2, and the constant 1. The derivative 
of a sum is the sum of the derivatives, so we differentiate with 
respect to x (which I will denote by an apostrophe, ')  :

     (x^2)' + (y^2)' + (1)' = (0)'

The first term is easy:

     (x^2)' = 2x

The derivative of a constant is 0, so 

     2x + (y^2)' + 0 = 0 

Since we are thinking of y as being a function of x, even though we 
don't know its formula, we can use the Chain Rule:

      d(y^2)    d(y^2)   dy
     -------- = ------ * ---
        dx        dy     dx


              =   2y * y'      (where I am now writing y' for dy/dx)

The first term on the right is just 2y, because we are differentiating 
y^2 WITH RESPECT TO y, so our equation becomes:

     2x + 2y*y' = 0

We now solve for y':

          2y*y' = -2x
  
                  -2x
             y' = ----
                   2y

                  -x
             y' = ---
                   y

Notice that y occurs in the answer. Remember how there were TWO 
functions defined by the circle equation:

     y1 = + sqrt(1 - x^2)

and

     y2 = - sqrt(1 - x^2)

We can differentiate these functions explicitly, and get

                -x                           -x
     y1' = -------------   and   y2' = --------------
           sqrt(1 - x^2)               -sqrt(1 - x^2)

or

           -x                           -x
     y1' = --              and    y2' = --
           y1                           y2

(Instead of canceling the - signs in the second derivative, I have 
left them in, to show that the single formula from implicit 
differentiation computes BOTH derivatives with one computation!)
   
For the second function, we would compute:

                   (y*e^y)' = (x)'

      y'*(e^y) + y * (e^y)' = 1   (I used the Product Rule on the 
                                   left)

      y'*e^y + y*(e^y * y') = 1   (since by the Chain Rule, 

                                    d(e^y)   d(e^y)   dy
                                    ------ = ------ * --
                                      dx       dy     dx

                                           = e^y * y'    )

Solving for y' gives

     y' * (e^y + y*e^y) = 1

       y' * e^y*(1 + y) = 1

                              1
                     y' = ---------
                          e^y*(1+y)

This answer again has the characteristic feature of implicit 
differentiation: y will appear in the answer.

I hope this helps explain the idea. If you have further questions, 
please write again.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/