But from (A ) ,
f
(
x
,
y
)
=
0.
∴
u
=
0
{\displaystyle f(x,y)=0.\therefore u=0}
d
u
d
x
=
0
;
{\displaystyle {\tfrac {du}{dx}}=0;}
(C )
∂
u
∂
x
+
∂
u
∂
y
d
y
d
x
=
0.
{\displaystyle {\frac {\partial u}{\partial x}}+{\frac {\partial u}{\partial y}}{\frac {dy}{dx}}=0.}
Solving for
d
y
d
x
,
{\displaystyle {\tfrac {dy}{dx}},}
[1]
(57 )
d
y
d
x
=
−
∂
u
∂
x
∂
u
∂
y
,
{\displaystyle {\frac {dy}{dx}}=-{\frac {\frac {\partial u}{\partial x}}{\frac {\partial u}{\partial y}}},}
a formula for differentiating implicit functions. This formula in the form (C ) is equivalent to the process employed in §62 , for differentiating implicit functions, and all the examples at the end of §63 may be solved by using formula (57 ) . Since
(D )
f
(
x
,
y
)
=
0
{\displaystyle f(x,\ y)=0}
for all admissible values of
x
{\displaystyle x}
y
{\displaystyle y}
(57 ) gives the relative time rates of change of
x
{\displaystyle x}
y
{\displaystyle y}
f
(
x
,
y
)
{\displaystyle f(x,y)}
Geometrically this means that the point
(
x
,
y
)
{\displaystyle (x,y)}
(D ) , and (57 ) determines the direction of its motion at any instant. Since
u
=
f
(
x
,
y
)
,
{\displaystyle u=f(x,\ y),}
we may write (57 ) in the form of
(57a )
d
y
d
x
=
−
∂
f
∂
x
∂
f
∂
y
{\displaystyle {\frac {dy}{dx}}=-{\frac {\frac {\partial f}{\partial x}}{\frac {\partial f}{\partial y}}}}
∂
f
∂
y
≠
0
{\displaystyle {\tfrac {\partial f}{\partial y}}\neq 0}
Illustrative Example 1.
x
2
y
4
+
sin
y
=
0
{\displaystyle x^{2}y^{4}+\sin y=0}
d
y
d
x
{\displaystyle {\tfrac {dy}{dx}}}
Solution. Let
f
(
x
,
y
)
=
x
2
y
4
+
sin
y
.
{\displaystyle f(x,y)=x^{2}y^{4}+\sin y.}
∂
f
∂
x
=
2
x
y
4
,
∂
f
∂
y
=
4
x
2
y
3
+
cos
y
.
{\displaystyle {\frac {\partial f}{\partial x}}=2xy^{4},{\frac {\partial f}{\partial y}}=4x^{2}y^{3}+\cos y.}
57a ),
d
y
d
x
=
−
2
x
y
4
4
x
2
y
3
+
cos
y
.
{\displaystyle {\frac {dy}{dx}}=-{\frac {2xy^{4}}{4x^{2}y^{3}+\cos y}}.}
Ans.
Illustrative Example 2.
x
{\displaystyle x}
x
=
3
{\displaystyle x=3}
y
{\displaystyle y}
y
=
1
{\displaystyle y=1}
2
x
y
2
−
3
x
2
y
{\displaystyle 2xy^{2}-3x^{2}y}
Solution. Let
f
(
x
,
y
)
=
2
x
y
2
−
3
x
2
y
{\displaystyle f(x,y)=2xy^{2}-3x^{2}y}
∂
f
∂
x
=
2
y
2
−
6
x
y
,
∂
f
∂
y
=
4
x
y
−
3
x
2
.
{\displaystyle {\frac {\partial f}{\partial x}}=2y^{2}-6xy,{\frac {\partial f}{\partial y}}=4xy-3x^{2}.}
Substituting in (57a ) ,
d
y
d
x
=
−
2
Y
2
−
6
x
y
4
x
y
−
3
x
2
,
{\displaystyle {\frac {dy}{dx}}=-{\frac {2Y^{2}-6xy}{4xy-3x^{2}}},}
d
y
d
t
d
x
d
t
=
−
2
y
2
−
6
x
y
4
x
y
−
3
x
2
.
{\displaystyle {\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}=-{\frac {2y^{2}-6xy}{4xy-3x^{2}}}.}
By (33 ),§94
But x = 3, y = 1,
d
x
d
t
=
2.
{\displaystyle {\tfrac {dx}{dt}}=2.}
d
y
d
t
=
−
2
2
15
,
{\displaystyle {\tfrac {dy}{dt}}=-2{\tfrac {2}{15}},}
Ans.
↑ It is assumed that
∂
u
∂
x
{\displaystyle {\tfrac {\partial u}{\partial x}}}
∂
u
∂
y
{\displaystyle {\tfrac {\partial u}{\partial y}}}
