Math 131.
Implicit Differentiation
Larson Section 2.5
So far we have dealt with differentiating explicitly√defined functions, that is, we are given the
expression defining the function, such as f (x) = 25 − x2 . However, equations can
√ define y
2
2
25 − x2 or
as a function
of
x,
for
example
the
equation
x
+
y
=
9
leads
to
functions
y
=
√
y = − 25 − x2 . Sometimes equation may define y as a function of x, but it may not be easy
or even possible to solve for y. Nevertheless, when y is differentiable, we can find the derivative
for y without solving the equation for y by using implicit differentiation, which will involve
dy
.
differentiating the equation with respect to x, and then solving for
dx
Before going into examples of implicit differentiation, let’s remind ourselves of a few basics.
dy
d
d
[y] =
, while
[x] = 1.
Remember, if y is a differentiable function of x, then
dx
dx
dx
Example 1. Suppose y is an unknown differentiable function of x. Evaluate the following
derivatives:
(a)
d 3
[y ]
dx
(d)
d
[sin(y)]
dx
(b)
(d)
d 3
[x ]
dx
d
[sin x]
dx
(c)
(e)
d 3 3
[x y ]
dx
d 3
[x sin(y)]
dx
Solution: (a) This is just a generalized power rule
d 3
dy
[y ] = 3y 2
dx
dx
[just think about what you would do if you knew, for example, y = x2 + 1]
(b) This is just the power rule,
d 3
[x ] = 3x2 .
dx
(c) Now you have a product of two functions that you cannot simplify further, so it requires
a product rule along with the answers from (a) and (b):
d 3
d
dy
d 3 3
dy
[x y ] =
[x ] · y 3 + x3 · [y 3 ] = 3x2 y 3 + x3 · 3y 2 ·
= 3x2 y 3 + 3x3 y 2
dx
dx
dx
dx
dx
(d) This involves the chain rule since y is a function of x, so
dy
d
[sin y] = (cos y)
dx
dx
(e) This is just the basic rule
d
[sin x] = cos x
dx
(f) This uses the product rule along with derivatives already done:
d 3
d
dy
d 3
[x sin(y)] =
[x ] sin(y) + x3 [sin(y)] = 3x2 sin y + x3 cos(y)
dx
dx
dx
dx
We will use techniques as in the previous example to differentiate both sides of an equation
dy
involving x and y with respect to x. Then we will solve the differentiated equation for
.
dx
dy
This process is called implicit differentiation. The expression for
will typically involve both
dx
x and y, to reduce it entirely to an equation in x would require knowing what the function y
is, and the point of implicit differentiation is to avoid needing to know the expression for the
function y.
Example 2. (a) Differentiate the equation x2 + y 2 = 25 implicitly to find
dy
dx
(b) Find the slope of the tangent line to this circle at the point (3, 4).
(c) Confirm your answer to (b), by solving the equation x2 + y 2 = 25 for y, and then taking
its derivative.
Solution: (a) First differentiate both sides of the equation with respect to x:
d
d 2
[x + y 2 ] =
[25]
dx
dx
Now solve the last equation for
2y
⇒ 2x + 2y
dy
=0
dx
dy
:
dx
dy
= −2x
dx
⇒
dy
2x
x
=−
=−
dx
2y
y
x
dy
3
dy
= − , and at the point (3, 4), x = 3 and y = 4 so
= − is the slope
dx
y
dx
4
of the tangent line.
√
(c) Solving x2 + y 2 = 25 for y we find y 2 = 25 − x2 and so
y
=
±
25 − x2 . At the point
√
(3, 4), y > 0 (it is on the top half of the circle), so y = 25 − x2 according to the chain
rule
1
x
dy
= (25 − x2 )−1/2 (−2x) = − √
dx
2
25 − x2
dy
3
3
when x = 3, we get
= −√
= − is the slope of the tangent line. Notice that this
dx
4
25 − 9
√
agrees with (b), in fact we can check the derivative agrees with (b) because y = 25 − x2
and so
dy
x
x
= −√
=−
2
dx
y
25 − x
(b) We know
A graph of the circle and tangent line is as follows
6 y
4
2
x
−6 −4 −2
−2
2
4
6
−4
−6
In the previous example, it was easy to solve for y and so implicit differentiation could have
been avoided. However, that is not the case in the next few examples.
Example 3. Find the slope of the tangent line to the Lemniscate 3(x2 + y 2 )2 = 100(x2 − y 2 )
at the point (4, 2).
Solution: Differenting the equation implicitly yields:
6(x2 + y 2 )(2x + 2yy 0 ) = 100(2x − 2yy 0 )
and dividing both sides by 4
3(x2 + y 2 )(x + yy 0 ) = 50(x − yy 0 )
Then rearranging this becomes
[3(x2 + y 2 )y + 50y]y 0 = 50x − 3x(x2 + y 2 )
and solving for y 0 yields
y0 =
50x − 3x(x2 + y 2 )
3y(x2 + y 2 ) + 50y
When x = 4 and y = 2, this means
y0 =
200 − 12(20)
40
2
=−
=− .
6(20) + 100
220
11
Then the tangent line has slope −2/11.
A very similar example to the previous is
Find the slope of the tangent line to the curve 2(x2 + y 2 )2 = 25(x2 − y 2 ) at
Example 4.
the point (3, 1).
Solution: First, differentiate implicitly
d
d
(2x4 + 2(2)x2 y 2 + 2y 4 ) =
(25x2 − 25y 2 )
dx
dx
and so
4(2)x3 + 4(2)xy 2 + 4(2)x2 yy 0 + 4(2)y 3 y 0 = 2(25)x − 2(25)yy 0
Dividing both sides by 2, and rearranging yields
(4x2 y + 4y 3 + 25y)y 0 = 25x − 4x3 − 4xy 2
dy
25x − 4x3 − 4xy 2
and so
= 2
.
dx
4x y + 4y 3 + 25y
When x = 3 and y = 1 we have
dy
−45
=
dx
65
Example 5. Find
dy
when 3x5 + 9x9 y − 3y 6 + sin(y) = 1.
dx
Solution: Differentiating with respect to x we find
15x4 + 81x8 y + 9x9 y 0 − 18y 5 y 0 + y 0 cos y = 0.
Next we wish to isolate y 0 on one side of the equation
(9x9 − 18y 5 + cos y)y 0 = −15x4 − 81x8 y
Now solve for y 0 to find
dy
−15x4 − 81x8 y
15x4 + 81x8 y
= 9
=
dx
9x − 18y 5 + cos y
18y 5 − 9x9 − cos y
The next few examples involve easier equations, but ask for tangent lines or where certain
tangent lines are on the graph.
Example 6. Find y 0 when y is implicitly defined by
p
p
√
√
4x + 2y + 3xy = 10 + 6
Then find the slope of the tangent line to this curve at the point (2, 1).
Solution: Differentiating implicitly we have
1
1
(4x + 2y)−1/2 (4 + 2y 0 ) + (3xy)−1/2 (3y + 3xy 0 ) = 0
2
2
and so
2
3x
√
+√
4x + 2y
3xy
4
3y
√
+√
4x + 2y
3xy
+
√3y
3xy
√3x
3xy
y =−
then
0
0
y =−
√ 4
4x+2y
√ 2
4x+2y
+
Plugging in x = 2 and y = 1 we have the slope of the tangent line at (2, 1) is
0
y =
√4
− 210
√
10
+
+
√3
6
√6
6
≈ −0.80781964
Example 7. Find the equation of the tangent line to the graph at the given point.
(x + 3)2 + (y − 2)2 = 26,
(−8, 3)
Solution: Using implicit differentiation, we find
d
d
[(x + 3)2 + (y − 2)2 ] =
[26]
dx
dx
and therefore, 2(y − 2)
⇒
2(x + 3) + 2(y − 2)
dy
=0
dx
dy
(x + 3)
dy
= −2(x + 3) and so
=−
.
dx
dx
(y − 2)
At the point (−8, 3), x = −8 and y = 3 and so
dy
−8 + 3
=−
=5
dx
3−2
The the tangent line has equation y − (3) = (5)(x − (−8)) which in slope-intercept form is
y = 5x + 43
A graph of the circle and tangent line is as follows
y
8
6
4
2
x
−12−10 −8 −6 −4 −2
−2
2
4
6
−4
Example 8. Find all points at which the graph of the following equation has a vertical or
horizontal tangent line.
9x2 + 4y 2 − 36x + 40y = −100
Solution: Using implicit differentiation, we find
d
d
[9x2 + 4y 2 − 36x + 40y] =
[−100]
dx
dx
solving the latter equation for
⇒
18x + 8y ·
dy
dy
− 36 + 40
=0
dx
dx
dy
dy
implies [8y + 40]
= −18x + 36 and so
dx
dx
dy
−18x + 36
=
.
dx
8y + 40
The horizontal tangent lines occur when
−18x + 36
=0
8y + 40
⇒
dy
= 0, and so
dx
−18x + 36 = 0
⇒
x=
36
=2
18
When x = 2, this implies 9(2)2 + 4y 2 − 36(2) + 40y = −100 and so
4y 2 + 40y + 64 = 0
This quadratic equation has solutions y = −8 and y = −2. The points where the tangent
lines are horizontal are then (2, −8) and (2, −2).
The graph of the equation is an ellipse, therefore it has vertical tangent lines where the
derivative is undefined, thus when 8y + 40 = 0, or y = −40/8 = −5. We plug y = −5 back
into the original equation to solve for x:
9x2 + 4(−5)2 − 36x + 40(−5) = −100
⇒
9x2 − 36x = 0
This quadratic equation has solutions x = 0 and x = 4. This means that the graph has
vertical tangent lines at the points (0, −5) and (4, −5).
Precalculus note. This problem could have been solved using precalculus: completing
the squares you can write the equation of the ellipse in standard form and then graph it.
In this case, the major axis is vertical, and so the horizontal tangent lines are at the end
points of the major axis and the vertical tangent lines are at the end points of the minor
axis.
Example 9.
Find the equations of the two lines tangent to the ellipse
x2 y 2
+
=1
16
9
that pass through the point (8, 0).
Solution: To find the equations of the two lines tangent to the ellipse
x2 y 2
+
=1
16
9
that pass through the point (8, 0), we first differentiate
x2 y 2
+
= 1 implicitly to find y 0 :
16
9
2x 9
9x
2x 2yy 0
+
= 0 and so y 0 = − ·
=−
16
9
16 2y
16y
A line tangent to the ellipse at the point (x0 , y0 ) passing through the point (8, 0) must have
slope
9x0
y0 − 0
=−
m=
x0 − 8
16y0
This implies 16y02 = −9x20 + (8)(9)x0 and so
(16)(9) − 9x20 = −9x20 + (8)(9)x0
√
16
3 3
Thus 16 = 8x0 and so x0 =
= 2 and then y0 = ±
. Thus, the tangent lines pass
2
√ 8 √ √
√
3 3
3 3
through the points 2, 3 2 3 and 2, − 3 2 3 , and have slopes − (3)(4)
and (3)(4)
respectively.
The equations of the tangent lines (which the reader can rewrite or simplify as desired)
will be
√
√
x
3y
x
3y
+
= 1 and
−
=1
8
6
8
6