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Unformatted text preview: Example Use implicit diﬀerentiation to ﬁnd y (x) if (x, y (x)) satisfy
x2 + 3xy + y 2 = 11,
and then ﬁnd the equation of the tangent line to this curve at the point (1 , 2). Solution: As a reminder that y depends on x, the above equation can be written
x2 + 3xy (x) + (y (x))2 = 11.
Diﬀerentiating both sides of the equation with respect to x,
d
d
(11),
x2 + 3xy (x) + (y (x))2 =
dx
dx
leads to
2x + 3x · y (x) + 3y (x) + 2y (x) · y (x) = 0, (1) where we have used the product rule to evaluate
d
(3x · y (x)) = (3x) · (y (x)) + (3x) · y (x) = 3xy (x) + 3y (x),
dx
and the Chain Rule to derive
d
(y (x))2 = 2y (x) · y (x).
dx
In order to solve equation (1) for y (x), collect all terms involving y (x) on the lefthand side
of the equation, and all other terms on the righthand side. That is,
3x · y (x) + 2y (x) · y (x) = −2x − 3y (x),
or, combining terms and replacing y (x) and y (x) by their simpler expressions y and y ,
respectively,
(3x + 2y )y = −2x − 3y,
(2)
an equation of the form
A(x, y )y = B (x, y ),
where
A(x, y ) = 3x + 2y, and B (x, y ) = −2x − 3y. Solving equation (2) for y results in
y= −2x − 3y
B (x, y )
=
.
A(x, y )
3x + 2y (3) The slope of the tangent line to the curve at the point (1, 2) is given by y in (3) where
the x and y values in (3) are given by 1 and 2, respectively. It follows that
y (1) = −2 · 1 − 3 · 2
8
=− .
3·1+2·2
7 So the equation of the tangent line to the curve
x2 + 3xy + y 2 = 11
at the point (1, 2) is given by
8
y − 2 = − (x − 1).
7 2 ...
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 Fall '10
 KIHYUNHYUN
 Calculus

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