Download MATH 1910 Implicit Differentiation x = − 4 8 2 0 x y + ⋅ = dy dx dy dx

MATH 1910
Implicit Differentiation
So far, all the relations between x and y have been explicitly solved for y as a function of x, as in
y = 4 − x2 .
However, there are relations you’ll encounter for which it is not possible to write y explicitly as a function of x.
y
An example of this would be 2 x + e = 4 y , impossible to solve for y in this equation.
This is an example of an implicit relation between x and y.
You’ll have to use implicit differentiation to find derivatives for implicit relations.
When you’re using implicit differentiation, you differentiate both x’s and y’s the same way. Since we’re assuming that y is a function
of x, anytime we differentiate a y we multiply it with " dy / dx " using the chain rule.
2
2
Ex. Differentiate 4 x + y = 16 .
SOLUTION:
Differentiating we get
Now, solve for
dy
dx
dy
8 x + 2 y ⋅ dx = 0
Æ
dy −8 x
4x
=
=−
.
dx 2 y
y
Æ
2
2
So, the implicit derivative of the ellipse 4 x + y = 16
is
2
dy
4x
=−
.
dx
y
2
Ex. Find the equation of the tangent line to the ellipse 4 x + y = 16 at the point ( 3, − 2) .
SOLUTION:
We’ve already got the implicit derivative for this relation:
dy
4x
=−
dx
y
The tangent line slope is found by plugging in the values from the point Æ
The equation of the tangent line is y − ( −2) = 2 3( x − 3)
3
2 2
4
Æ
dy
4 3
=−
=2 3
dx
−2
y = 2 3 ⋅ x −8
Ex. Differentiate y + 3x y + 5x = 12
SOLUTION: Differentiate left to right and apply the product rule to the middle term:
3y 2
Now, solve for
dy
dy
+ 6 xy 2 + 6 x 2 y + 20 x 3 = 0
dx
dx
dy
:
dx
dy
dy
+ 6x 2 y
= −6 xy 2 − 20 x 3
dx
dx
dy
3y 2 + 6x 2 y
= −6 xy 2 − 20 x 3
dx
2
dy −6 xy − 20 x 3
=
3y 2 + 6x 2 y
dx
3y 2
c
h
MATH 1910
Implicit Differentiation
Ex. Differentiate x sin y + cos 2 y = cos y .
SOLUTION: Differentiate left to right and apply the product rule to the first term:
1⋅ sin y + x ⋅ cos y ⋅
Now, solve for
dy
dy
dy
+ ( −2 sin 2 y ⋅ ) = − sin y ⋅
dx
dx
dx
Å product and chain rule involved
dy
:
dx
dy
dy
dy
dy
− 2 sin 2 y + sin y ⋅ = − sin y Å collect all the
terms on the left
dx
dx
dx
dx
dy
dy
x cos y − 2 sin 2 y + sin y
= − sin y
Å factor out the
on the left
dx
dx
− sin y
dy
dy
=
Å divide to solve for
dx x cos y − 2 sin 2 y + sin y
dx
x cos y
b
g
Ex. Show that the given curves are orthogonal (have perpendicular derivatives at their points of intersection)
x 2 − y 2 = 5 , 4 x 2 + 9 y 2 = 72
SOLUTION: Differentiate both curves implicitly
For
x2 − y2 = 5: 2x − 2 y
dy
dy
= 0 Æ −2 y
= −2 x Æ
dx
dx
For 4 x + 9 y = 72 : 8 x + 18 y
2
2
dy x
=
dx y
dy
dy
4x
dy
= 0 Æ 18 y
= −8 x Æ
=−
9y
dx
dx
dx
Now we need to find where the curves intersect.
x 2 = y 2 + 5 into the other equation 4 x 2 + 9 y 2 = 72 :
4 x 2 + 9 y 2 = 72 Æ 4( y 2 + 5) + 9 y 2 = 72 Æ 13 y 2 = 52 Æ y 2 = 4 Æ y = ±2
2
2
2
2
Now, since x = y + 5 , we get the x solutions: x = ( ±2) + 5 = 9 Æ x = ±3
These two curves intersect at 4 points: (3, 2), ( −3, 2), (3, − 2), (−3, − 2)
Solving the system would be best done by substitution, make the substitution
Plugging each point pair into the derivatives gives us the following slopes:
x2 − y2 = 5
4 x 2 + 9 y 2 = 72
(3,2)
dy 3
=
dx 2
( −3,2)
dy
3
=−
2
dx
(3,−2)
dy
3
=−
2
dx
( −3,−2)
dy 3
=
dx 2
dy
2
=−
dx
3
dy 2
=
dx 3
dy 2
=
dx 3
dy
2
=−
dx
3
So these curves are perpendicular, or orthogonal, at their points of intersection because their derivative values at those points are
opposite reciprocals.
MATH 1910
Implicit Differentiation
Derivative of the Natural Logarithm function
This derivative will not be used until section 3.7 but we will use implicit differentiation to derive it.
Start with
y = ln x , and rewrite it as e y = x
Now, use implicit differentiation on
dy
=1
dx
dy 1
=
dx e y
1
dy
= ln x
dx e
ey
ey = x
Å solve for
dy
dx
Å now make the substitution
Æ
dy 1
=
dx x
So our final function derivative for this chapter is
The chain rule version of this derivative is
y = ln x
Å the cancellation property causes
d
1
(ln x ) =
dx
x
d
1
f ′( x )
[ln( f ( x ))] =
⋅ f ′( x ) =
dx
f ( x)
f ( x)
e ln x = x