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```17. IMPLICIT DIFFERENTIATION
53
17. Implicit Differentiation
In the last several sections, we computed the derivatives of many
diﬀerent functions. Although these functions were diﬀerent, they had
one important feature in common. They were explicitly given. That
is, they were given by a rule that directly described how f (x) = y is
obtained from x.
17.1. Tangent Lines to Implicitly Deﬁned Curves. Sometimes we have to
deal with curves that are given by a diﬀerent kind of rule. Consider
the curve given by the equation
(3.12)
x3 + y 3 = 4xy.
Let us say that we want to compute the slope of the tangent line
to this curve at the point (2, 2). If we could express y as a function
of x, we could simply take the derivative of that function at x = 2.
However, it is not clear how to write y explicitly in terms of x, even if
(3.12) implicitly describes this dependence.
It is in these situations that we resort to implicit diﬀerentiation.
Keep in mind that we do not need to explicitly know how y depends
on x, that is, we do not need an explicit expression for the function
y(x); we only need to know the derivative dy/dx of that function at
x = 2.
Consider Equation (3.12), and diﬀerentiate both sides with respect
to the variable x to get
d
d 3
x + y3 =
(4xy).
dx
dx
Now recall that y = y(x) is a function of x. So, when computing
(d/dx)y 3 on the left-hand side, we need to use the chain rule. On the
right-hand side, we need to use the product rule. Using these rules,
we get
dy
dy
= 4y + 4x .
3x2 + 3y 2
dx
dx
Expressing dy/dx from this equation, we get
dy
(4y − 3x2 )
=
.
dx
(3y 2 − 4x)
At the point (2, 2), the right-hand side is −4/4 = −1, so the slope of
the tangent line at (2, 2) is −1.
Note that the fact that the tangent line at (2, 2) has slope −1 makes
(intuitively) perfect sense, since the curve in question is symmetric in
54
3. RULES OF DIFFERENTIATION
x and y. That is, if (x, y) is on the curve, then (y, x) is also on the
curve.
17.2. Derivatives of Inverse Trigonometric Functions. One place where
implicit diﬀerentiation is a very powerful tool is in the computation of
the derivatives of inverse trigonometric functions. Recall that tan−1 x =
y is the function that is the inverse of the restriction of the function
tan x to the interval (−π/2, π/2). That is, if
tan−1 (x) = y,
then
(3.13)
x = tan y,
where y ∈ (−π/2, π/2).
Our goal is to determine
d
dy
tan−1 (x) =
.
dx
dx
To that end, let us take the derivative of both sides of (3.13) with
respect to x. Recalling that
d
tan z = sec2 z
dz
and y = y(x),
we get
dy
.
dx
Solving for dy/dx and recalling the identity sec2 z = 1 + tan2 z, we
obtain
1
1
1
dy
=
=
=
.
2
2
dx
sec y
1 + tan y
1 + x2
1 = sec2 y ·
In other words, we proved the suprisingly simple formula
(tan−1 x) =
1
.
1 + x2
This formula is interesting for two reasons. First, it is surprisingly simple. Second, it does not even contain trigonometric functions. Imagine
trying to get this result without implicit diﬀerentiation, using just the
deﬁnition of derivatives.
You will be asked to compute the derivatives of the other inverse
trigonometric functions in the exercises.
```