3.5 Implicit Differentiation EXAMPLE Suppose we want to the slope of the tangent to the graph of 3x 2 + 2y −1 = x 3 + y + 4 at the point (1, 3). This means that we need to differentiate the equation, and find the value of the derivative at x = 1. To find the derivative, we would probably first rewrite the equation, solving for y in terms of x, so that we can express y explicitly as a function of x: y = x 3 − 3x 2 + 5 Now, we find the derivative and evaluate it at x = 1. dy
= 3x 2 − 6x
dx
dy
= −3
dx x=1
However, we could also have arrived at this result without having to write y explicitly as a function of x: 3x 2 + 2y −1 = x 3 + y + 4 Differentiate both sides with respect to x: d
d 3
3x 2 + 2y −1 =
x + y + 4 dx
dx
dy
dy
6x + 2 = 3x 2 + dx
dx
Solve for dy/dx: dy dy
2 − = 3x 2 − 6x
dx dx
[
]
[
]
dy
= 3x 2 − 6x
dx
This second approach is an example of implicit differentiation. Rather than solving the equation explicitly for y in terms of x, and then differentiating with respect to x, we differentiate each term of the original equation, treating y as an implicit function of x. Correct use of implicit differentiation depends upon the chain rule. Recall the Chain Rule: d
f ( g( x))) = f ′( g( x)) ⋅ g′( x) (
dx
This can also be expressed as follows: Suppose, in the expression above, that g(x) = u (that is, suppose u is a function of x) Then d
f (u)) = f ′(u) ⋅ u′ (
dx
or d
du
f (u)) = f ′(u) ⋅ (
dx
dx
EXAMPLE Suppose we want to find the slope of the tangent line to the graph of the equation x 2 + 2y2 = 9 at the point (–1, 2). Note that we cannot write y explicitly as a function of x. If we solve the equation for y in terms of x we get two solutions: 9 − x2
9 − x2
y=
y=−
2
2
This corresponds to the fact that the equation x 2 + 2y2 = 9 is not a function (its graph fails the vertical line test). However, it is possible to differentiate this equation with respect to x by treating y is an implicit function of x. This means that we differentiate each term, using the usual rules for differentiation, but that any term involving y will also involve the chain rule, as follows. Differentiate both sides of the equation, with respect to x: d 2
d
x + 2y2 = (9) dx
dx
d 2
d
x +
2y2 = 0 dx
dx
Differentiate both terms on the left side, treating y as a function of x: dy
2x + 4 y = 0 dx
dy
Now, we solve for : dx
(
)
( )
( )
Finally, find the value of −
dy
when x = –1, y = 2 dx
x
= 2y x=−1
y=2
We use implicit differentiation to differentiate an equation in which it is not possible or practical to express one variable explicitly as a function of the other variable. EXAMPLE Find dy/dx by implicit differentiation if 1+ x = sin xy2 ( )
EXAMPLE Find use implicit differentiation to dy/dx for x 2 + 2xy − y 2 = xy 2 + x + y EXAMPLE Use implicit differentiation to find dy/dx for y cos x = 5x 2 + 4 y 2 EXAMPLE Use implicit differentiation to find dy/dx for e x y = 8x − y Derivatives of Inverse Trig Functions The following formulas will be provided for you if they are needed for a test or quiz question. They are presented in section 3.5 because they are proven through implicit differentiation. d
1
d
1
arcsin(x) =
arccos(x)
=
−
2
2
dx
dx
1− x
1− x
d
1
d
1
arctan(x) =
arccot(x)
=
−
dx
1+ x 2
dx
1+ x 2
d
1
d
1
arcsec(x) =
arccsc(x)
=
−
2
2
dx
dx
x x −1
x x −1
EXAMPLE Use implicit differentiation to prove d
1
arctan(x) =
2
dx
1+ x
EXAMPLE d
1
Use arctan(x) =
to find the derivative of dx
1+ x 2
(
)
y = arctan 2 x + sec x Do not simplify.