Differentiation of implicit functions 323
i.e. 2 x+ 2 ydy
dx= 0Hencedy
dx=−2 x
2 y=−x
ySincex^2 +y^2 =25, whenx=4,y=
√
( 25 − 42 )=± 3Thus whenx=4andy=±3,
dy
dx
=−
4
± 3=±
4
3x^2 +y^2 =25 is the equation of a circle, centre at the
origin and radius 5, as shown in Fig. 30.1. Atx=4, the
two gradients are shown.
y
53(^045) x
23
25
25
Gradient
3524
Gradient
54
3
x^21 y^2525
Figure 30.1
Above,x^2 +y^2 =25 was differentiated implicitly;
actually, the equation could be transposed to
y=
√
( 25 −x^2 )and differentiated using the function of
a function rule. This gives
dy
dx
1
2
( 25 −x^2 )
− 1
(^2) (− 2 x)=−
x
√
( 25 −x^2 )
and whenx=4,
dy
dx
=−
4
√
( 25 − 42 )
=±
4
3
as obtained
above.
Problem 8.
(a) Find
dy
dx
in terms ofxandygiven
4 x^2 + 2 xy^3 − 5 y^2 =0.
(b) Evaluate
dy
dx
whenx=1andy=2.
(a) Differentiating each term in turn with respect tox
gives:
d
dx
( 4 x^2 )+
d
dx
( 2 xy^3 )−
d
dx
( 5 y^2 )=
d
dx
( 0 )
i.e. 8x+
[
( 2 x)
(
3 y^2
dy
dx
)
+(y^3 )( 2 )
]
− 10 y
dy
dx
= 0
i.e. 8x+ 6 xy^2
dy
dx
2 y^3 − 10 y
dy
dx
= 0
Rearranging gives:
8 x+ 2 y^3 =( 10 y− 6 xy^2 )
dy
dx
and
dy
dx
8 x+ 2 y^3
10 y− 6 xy^2
4 x+y^3
y(5− 3 xy)
(b) Whenx=1andy=2,
dy
dx
4 ( 1 )+( 2 )^3
2[5−( 3 )( 1 )( 2 )]
12
− 2
=− 6
Problem 9. Find the gradients of the tangents
drawntothecirclex^2 +y^2 − 2 x− 2 y=3atx=2.
The gradient of the tangent is given by
dy
dx
Differentiatingeach term in turn with respect toxgives:
d
dx
(x^2 )+
d
dx
(y^2 )−
d
dx
( 2 x)−
d
dx
( 2 y)=
d
dx
( 3 )
i.e. 2 x+ 2 y
dy
dx
− 2 − 2
dy
dx
= 0
Hence ( 2 y− 2 )
dy
dx
= 2 − 2 x,
from which
dy
dx
2 − 2 x
2 y− 2
1 −x
y− 1
The value ofy whenx=2 is determined from the
original equation.
Hence( 2 )^2 +y^2 − 2 ( 2 )− 2 y= 3
i.e. 4 +y^2 − 4 − 2 y= 3
or y^2 − 2 y− 3 = 0
Factorizing gives: (y+ 1 )(y− 3 )=0, from which
y=−1ory=3.
Whenx=2andy=−1,
dy
dx
1 −x
y− 1
1 − 2
− 1 − 1
− 1
− 2
1
2