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2.7. DERIVATIVES OF FUNCTIONS GIVEN IMPLICITLY
point on the circle? By viewing y as an implicit 10 function of x, we essentially think of y as
some function whose formula f (x) is unknown, but which we can differentiate. Just as y
represents an unknown formula, so too its derivative with respect to x,
dy
dx , will be (at least
temporarily) unknown.
Consider the equation x 2 + y 2 = 16 and view y as an unknown differentiable function
of x. Differentiating both sides of the equation with respect to x, we have
d
dx
x
2 + y
2
=
d
dx
[16] .
On the right, the derivative of the constant 16 is 0, and on the left we can apply the sum
rule, so it follows that
d
dx
x
2
+
d
dx
y
2
= 0.
Next, it is essential that we recognize the different roles being played by x and y. Since x is
the independent variable, it is the variable with respect to which we are differentiating, and
thus
d
dx
x 2
= 2x. But y is the dependent variable and y is an implicit function of x. Thus,
when we want to compute
d
dx [y 2 ] it is identical to the situation in Preview Activity 2.7
where we computed
d
dx [ f (x) 2 ]. In both situations, we have an unknown function being
squared, and we seek the derivative of the result. This requires the chain rule, by which
we find that
d
dx [y 2 ] = 2y 1 dy
dx . Therefore, continuing our work in differentiating both sides
of x 2 + y 2 = 16, we now have that
2x + 2y
dy
dx
= 0.
Since our goal is to find an expression for
dy
dx , we solve this most recent equation for
dy
dx .
Subtracting 2x from both sides and dividing by 2y,
dy
dx
= −
2x
2y
= −
x
y
.
There are several important things to observe about the result that
dy
dx = −
x
y . First,
this expression for the derivative involves both x and y. It makes sense that this should be
the case, since for each value of x between −4 and 4, there are two corresponding points
on the circle, and the slope of the tangent line is different at each of these points. Second,
this formula is entirely consistent with our understanding of circles. If we consider the
radius from the origin to the point (a, b), the slope of this line segment is m r =
b
a . The
tangent line to the circle at (a, b) will be perpendicular to the radius, and thus have slope
m t = −
a
b , as shown in Figure 2.15. Finally, the slope of the tangent line is zero at (0, 4) and
10 Essentially the idea of an implicit function is that it can be broken into pieces where each piece can be
viewed as an explicit function of x, and the combination of those pieces constitutes the full implicit function.
For the circle, we could choose to take the top half as one explicit function of x, and the bottom half as
another.
2.7. DERIVATIVES OF FUNCTIONS GIVEN IMPLICITLY
point on the circle? By viewing y as an implicit 10 function of x, we essentially think of y as
some function whose formula f (x) is unknown, but which we can differentiate. Just as y
represents an unknown formula, so too its derivative with respect to x,
dy
dx , will be (at least
temporarily) unknown.
Consider the equation x 2 + y 2 = 16 and view y as an unknown differentiable function
of x. Differentiating both sides of the equation with respect to x, we have
d
dx
x
2 + y
2
=
d
dx
[16] .
On the right, the derivative of the constant 16 is 0, and on the left we can apply the sum
rule, so it follows that
d
dx
x
2
+
d
dx
y
2
= 0.
Next, it is essential that we recognize the different roles being played by x and y. Since x is
the independent variable, it is the variable with respect to which we are differentiating, and
thus
d
dx
x 2
= 2x. But y is the dependent variable and y is an implicit function of x. Thus,
when we want to compute
d
dx [y 2 ] it is identical to the situation in Preview Activity 2.7
where we computed
d
dx [ f (x) 2 ]. In both situations, we have an unknown function being
squared, and we seek the derivative of the result. This requires the chain rule, by which
we find that
d
dx [y 2 ] = 2y 1 dy
dx . Therefore, continuing our work in differentiating both sides
of x 2 + y 2 = 16, we now have that
2x + 2y
dy
dx
= 0.
Since our goal is to find an expression for
dy
dx , we solve this most recent equation for
dy
dx .
Subtracting 2x from both sides and dividing by 2y,
dy
dx
= −
2x
2y
= −
x
y
.
There are several important things to observe about the result that
dy
dx = −
x
y . First,
this expression for the derivative involves both x and y. It makes sense that this should be
the case, since for each value of x between −4 and 4, there are two corresponding points
on the circle, and the slope of the tangent line is different at each of these points. Second,
this formula is entirely consistent with our understanding of circles. If we consider the
radius from the origin to the point (a, b), the slope of this line segment is m r =
b
a . The
tangent line to the circle at (a, b) will be perpendicular to the radius, and thus have slope
m t = −
a
b , as shown in Figure 2.15. Finally, the slope of the tangent line is zero at (0, 4) and
10 Essentially the idea of an implicit function is that it can be broken into pieces where each piece can be
viewed as an explicit function of x, and the combination of those pieces constitutes the full implicit function.
For the circle, we could choose to take the top half as one explicit function of x, and the bottom half as
another.