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### Implicit vs. Explicit Differentiation

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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,

- Doctor Fenton, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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