Download x2 +y2 = 25. We know the graph of this to be a

Calculus and Analytic Geometry I
Implicit Differentiation
Consider the equation: x2 + y 2 = 25. We know the graph of this to be a circle of radius 5 centered at the origin,
and we know it is not a function. However, if we understand that some ’cutting’ might be done, we can view
x2 + y 2 = 25 as containing several functions. Note
5
-10
-5
5
0
5
10
=
-10
-5
5
0
-5
5
10
&
-10
-5
-5
5
10
-5
√
y = + 25 − x2
x2 + y 2 = 25
0
√
y = − 25 − x2
So there are really functions sitting inside relationships even though we have not (or can not) solve those
relationships for y in terms of x. We call functions of this type implicit functions, and this chapter discusses
derivatives of such functions. The process of finding
dy
dx
when y is an implicit function of x is called implicit
differentiation.
Before embarking upon implicit differentiation lets compute the derivatives of each of the following, with respect
to the appropriate variable, as directed. Consider u, s, and y to represent some expression which is a function
of x.
1.
d 3
(x)
dx
2.
d 2
(x + 1)3
dx
3.
d 2
(x + cos(x))3
dx
4.
d 3
(u)
dx
5.
d 3
(s)
dx
6.
d 3
(y)
dx
Suppose we want to find the slopes of the two tangent lines
pictured at right—the lines tangent to the circle x2 + y 2 = 25
at the points (4, 3) and (4, −3).
Taking the derivative of each side of the equation x2 +y 2 = 25,
we get:
2
d
2
=
dx x + y
d
dx
[25]
Note that since we have a derivative of a sum, we can take
each derivative separately.
2
2
d
d
+ dx
y =0
dx x
Hence, noting that the chain rule is necessary for the derivative of y 2 with respect to x (considering y to be an
implicit function of x), we have
dy
2x + 2y dx
=0
and solving for x we get:
at the point (4, 3) is
is
dy
dx
4
= − −3
= 4/3
dy
dx
dy
dx
= − xy . From this we see that the slope of the line tangent to the circle x2 + y 2 = 25
= − 43 and the slope of the line tangent to the circle x2 + y 2 = 25 at the point (4, −3)
Example 1A: Compute the implicit derivative of x2 y 2 − sin(xy) = −1.
Solution: Taking the derivative of each side of the equation, we get:
2 2
d
d
dx x y − sin(xy) = dx [−1]
Note that since we have a derivative of a difference, we can take each derivative separately.
2 2
d
d
[sin(xy)] = 0
− dx
dx x y
Now we have a product and an composition, so we use the product rule and chain rule (respectively) to get
2
d
d
2x · y 2 + x2 dx
y − cos(xy) dx
[xy] = 0
Since y is an implicit function of x, we have another chain rule (and a product rule) to get
2x · y 2 + x2 · 2y ·
dy
dx
− cos(xy) · (1 · y + x ·
dy
dx )
=0
that is
2xy 2 + 2x2 y ·
dy
dx
− y cos(xy) − x cos(xy) ·
Collecting terms with
2x2 y ·
dy
dx
− x cos(xy) ·
(2x2 y − x cos(xy)) ·
dy
dx
dy
dx ,
dy
dx
=0
dy
dx
to solve for
dy
dx
= 0 − 2xy 2 + y cos(xy) and so
we get:
= 0 − 2xy 2 + y cos(xy)
and finally:
dy
dx
=
2xy 2 +y cos(xy)
2x2 y−x cos(xy)
Example 1B: Compute the implicit derivative of xy 3 − 3 tan(xy) = 11.
Solution:
Example 1C: Find an equation for the line tangent to the graph of
point (0, 1).
2x2 y 3 − 2exy + 2y cos(x) = 2
at the