CHAPTER 3. RULES FOR DERIVATIVES
3.5
70
Implicit derivatives
Example 1. Find the equation of the tangent line at x = 1/2 on the circle x2 +y 2 =
1, by solving explicitly for y. Graph the result.
√
Solution. y = 1 − x2 .
1
y � = (1 − x2 )−1/2 · 2x.
2
√
1
1
1
1
m = y � (1/2) = (1−(1/2)2 )−1/2 ·2· = �
·1 = �
= −1/ 3.
2
2
2 1 − 1/4
2 3/4
�
√ �
1 3
Example 2. Find the slope at
,
on the circle x2 + y 2 = 1, by working
2 2
implicitly: remember, y is a function of x.
Solution. We take the derivative of both sides of x2 + y 2 = 1,
d 2
d
(x + y 2 ) =
1
dx
dx
d
2x + y 2 = 0
dx
d 2
To figure out
y we use the chain rule (after all, y is somehow a function
dx
of x, so we have one function, inside another function, the outside one is the
squaring function).
�
d
2
=2
·
dx
We plug y into this,
�
d 2
y = 2 y · y = 2y · y �
dx
�
� �
Thus, the above equation becomes 2x + 2y · y � = 0
2y · y � = −2x
x
y� = −
y
� √ �
1 3
At the point
,
we have
2 2
√
1/2
y� = − √
= −1/ 3
3/2
Note, the implicit formula shows us something nice about the slope of a point
y
on the circle: if the point is (x, y), then the slope of the radius will be . Note
x
that each tangent line is perpendicular to the radius at that point. Therefore,
1
the slope of a tangent line will be −
, which is what we found.
y/x
Example 3. Find the equation of the tangent line at x = 2/3, y = 4/3 for the
shape defined by x3 + y 3 = 3xy (shown below). Add the graph of the tangent line
to the graph of x3 + y 3 = 3xy shown below.
CHAPTER 3. RULES FOR DERIVATIVES
d 3
d
(x + y 3 ) =
(3xy)
dx
dx
d
d
3x2 + y 3 = 3 (xy)
dx
dx
d 3
For
y we use the chain rule:
dx
Solution.
�
d 3
y = 3 y 2 · y = 3y 2 · y �
dx
For
d
(xy) we use the product rule
dx
x� · y + x · y � = y + x · y �
Thus we have
3x2 + 3y 2 · y � = 3(y + x · y � )
Now we solve for y � :
3x2 + 3y 2 · y � = 3y + 3x · y �
3y 2 · y � − 3x · y � = 3y − 3x2
(3y 2 − 3x)y � = 3y − 3x2
3y − 3x2
y� = 2
3y − 3x
y − x2
y� = 2
y −x
Plug in x = 2/3 and y = 4/3 to get m = 4/5.
71
CHAPTER 3. RULES FOR DERIVATIVES
72
graphics/folium_of_descartes_with_tangent-eps-converted-to.pdf
Example 4. Find y � for ey cos(x) = 1 + sin(x/y)
Solution.
d y
d
e cos(x) =
(1 + sin(x/y))
dx
dx
(ey )� (cos(x)) + (ey )(cos(x))� = 0 + (sin(x/y))�
(x)� (y) − (x)(y)�
ey · y � cos(x) − ey · sin(x) = cos(x/y) ·
(y)2
y − xy �
ey · y � cos(x) − ey · sin(x) = cos(x/y) ·
y2
y
xy �
ey · y � cos(x) − ey · sin(x) = cos(x/y) 2 − cos(x/y) 2
y
y
�
xy
1
ey · y � cos(x) + cos(x/y) 2 = cos(x/y) + ey · sin(x)
y
y
�
�
1
x
ey cos(x) + cos(x/y) 2 y � = cos(x/y) + ey · sin(x)
y
y
cos(x/y) y1 + ey · sin(x)
�
�
�
y =
x
y
e cos(x) + cos(x/y) y2
Inverse trig functions
Example 5. Suppose sin(θ) =
Solution.
follows
From sin(θ) =
5
(and 0 < θ < π/2). Find cos(θ) and tan(θ).
7
5
we can draw a triangle and label two sides as
7
CHAPTER 3. RULES FOR DERIVATIVES
7
73
5
θ
?
Then the missing side is
5
tan(θ) = √ .
6 2
√
72 − 52 =
√
√
√
24 = 2 6 and so cos(θ) = 6 2/7,
Example 6. Suppose sin(y) = x, find cos(y) and tan(y).
Solution.
follows
From sin(y) = x we can draw a triangle and label two sides as
1
x
y
?
Then the missing side is
x
√
.
1 − x2
√
1 − x2 and so cos(y) =
√
1 − x2 and tan(y) =
Proof. Prove the sine inverse rule by turning it into an implicit derivative.
y = sin−1 (x)
sin(y) = x
d
d
sin(y) =
x
dx
dx
cos(y) · y � = 1
1
y� =
cos(y)
⇒ cos(y) =
1
y� = √
1 − x2
sin(y) = x
Example 7. Find the derivative of
�
√
1 − x2
Example 6
1 − x2 arctan(πx + 7)
Solution. Note that arctan is another name for tan−1 .
√
√
d√
1 − x2 arctan(πx + 7) = ( 1 − x2 )� · arctan(πx + 7) + 1 − x2 · (arctan(πx + 7))�
dx
product rule
CHAPTER 3. RULES FOR DERIVATIVES
74
√
1
1
= √
(−2x) · arctan(πx + 7) + 1 − x2 ·
π
2
1 + (πx + 7)2
2 1−x
chain rule