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defines y as several functions of x.Whenwesaythatf is a function
defined implicitly by (2), we mean that the equation x
3
+(f(x))
3
=
6xf(x) is true for all values of x in the domain of f.
Fortunately, it is not necessary to solve an equation for y in terms
of x in order to find the derivative of y. Instead we can use the method
of implicit differentiation. This consists of differentiating both sides
of the equation with respect to x and then solving the resulting
equation for y
.
Example 12
If x
2
+ y
2
=25, find
dy
dx
.
Differentiating both sides of the equation
d
dx
(x
2
+ y
2
)=
d
dx
(25).
Remembering that y is a function of x and using the Chain Rule, we
have
(x
2
)
+(y
2
)
=0, 2x +2yy
=0.
Now we solve for y
: y
= −
x
y
.
Example 13
Find the equation of the tangent to the circle x
2
+ y
2
=25at
the point (3, −4).
S o l u t i o n 1. At the point (3, −4) we have x =3and y = −4,
so y
=3/4. An equation of the tangent to the circle at (3, −4) is
therefore
y +4=
3
4
(x − 3) or 3x − 4y =25.
S o l u t i o n 2. Solving the equation x
2
+ y
2
=25,wegety =
±
√
25 − x
2
. The point (3, −4) lies on the lower semicircle
y = −
√
25 − x
2
and so we consider the function f(x)=−
√
25 − x
2
.
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