MATH 12002 - CALCULUS I
§2.6: Implicit Differentiation
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
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Implicit Differentiation
A relation between x and y represents a graph in the xy -plane;
i.e., the set of points (a, b) whose coordinates x = a, y = b satisfy the
relation. If the relation is not a function, we will not be able to write y as
a single function of x.
However, if the graph is a smooth curve, it will still have tangent lines and
we should be able to find the slopes of the tangent lines, even though y is
not of the form y = f (x) and the slope cannot be written as f 0 (a).
We will see how to find this slope (the derivative of y with respect to x)
implicitly, but first let’s consider an example where we can find the slope
explicitly.
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Implicit Differentiation
Example
2
2
Find the slope
√ of the line tangent to the ellipse 9x + 4y = 36 at the
point (1, − 3 2 3 ).
Solution 1
We can solve for y in terms of x: since 9x 2 + 4y 2 = 36, we have
4y 2 = 36 − 9x 2
y2 =
y
=
2
1
4 (36 − 9x )
p
± 21 36 − 9x 2 .
√
Here, y = 21 36 − 9x 2 represents the top half of the ellipse (y > 0) and
√
y = − 12 36 − 9x 2 represents the bottom half (y 6 0).
[Continued →]
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Implicit Differentiation
Solution 1 [continued]
√
Since our point, (1, − 3 2 3 ), is on the bottom half of the ellipse, the slope
√
is f 0 (1), where f (x) = − 12 36 − 9x 2 . We have
f 0 (x) = − 21 · 21 (36 − 9x 2 )−1/2 · (−18x)
9x
√
,
=
2 36 − 9x 2
and so the slope is
√
9
9
3·3
3
√ =
f (1) = p
= √
= √ =
.
2
2
2 36 − 9
2 27
2·3 3
2 36 − 9(1 )
0
9(1)
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Implicit Differentiation
Solution 2
We now find the slope implicitly.
We know that the top and bottom halves of the ellipse are functions.
Instead of solving for the function explicitly, let’s just say y = f (x), so that
9x 2 + 4[f (x)]2 = 36.
Now take derivatives of both sides of the equation with respect to x:
18x + 8[f (x)]1 · f 0 (x) = 0.
dy
dx ” and 18x + 8y
then dy
dx evaluated at
Since “f (x) = y ” this means “f 0 (x) =
and we get
dy
dx
=
9x
− 4y
.
The slope is
·
dy
dx
= 0,
√
(1, − 3 2 3 ), i.e.,
√
dy 9(1)
3·3
3
√
√ =
=−
=
.
√
3 3
dx (1,− 3 3 )
2
2
·
3
3
4(−
)
2
2
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Examples
1
Let x 2 + x 3 y 5 + y −3 = 4 and find y 0 =
respect to x.
dy
dx ,
the derivative of y with
Instead of writing y as f (x), we will just treat y as some function of x,
so that y 5 , for example, is a composite function with inside function y .
First, take the derivative with respect to x of both sides of the equation
x 2 + x 3 y 5 + y −3 = 4, to obtain
2x + [3x 2 y 5 + x 3 · 5y 4 y 0 ] − 3y −4 y 0 = 0.
Now solve for y 0 in terms of x and y :
2x + 3x 2 y 5 = −5x 3 y 4 y 0 + 3y −4 y 0
2x + 3x 2 y 5 = (−5x 3 y 4 + 3y −4 )y 0
y0 =
D.L. White (Kent State University)
2x + 3x 2 y 5
.
−5x 3 y 4 + 3y −4
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Examples
2
Let 2y 3 = tan(x 2 + y 2 ) and find y 0 =
respect to x.
dy
dx ,
the derivative of y with
First, take the derivative with respect to x of both sides of the equation
2y 3 = tan(x 2 + y 2 ) to obtain
6y 2 y 0 = [sec2 (x 2 + y 2 )](2x + 2yy 0 ).
Now solve for y 0 in terms of x and y :
6y 2 y 0 = [sec2 (x 2 + y 2 )] · 2x + [sec2 (x 2 + y 2 )] · 2yy 0
6y 2 y 0 − [sec2 (x 2 + y 2 )] · 2yy 0 = [sec2 (x 2 + y 2 )] · 2x
6y 2 − [sec2 (x 2 + y 2 )] · 2y y 0 = [sec2 (x 2 + y 2 )] · 2x
y0 =
D.L. White (Kent State University)
[sec2 (x 2 + y 2 )] · 2x
.
6y 2 − [sec2 (x 2 + y 2 )] · 2y
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