Part II: The Answers
274
Answers
401–500
Next, find the second derivative:
d y
dx
x
y
dy
dx
y
x
x
2
2
1 2
1 2
1 2
1 2
1 2
2
1
2
1
2
=
−
−−
−
−





 ( ) ( )
( )
Substituting in the value of dy
dx
y
x
=
−
1 2
1 2
gives you
d y
dx
x y
y
x
y
x
x
y
x
x
x
y
2
2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1
1
2
2
1
2 2
=
−
−
+
=
+
=
+
−






2 2
3 2
2x
And now using x
y
+
=1 gives you the answer:
d y
dx
x
2
2
32
1
2
=
418.
y = –x + 2
You know a point on the tangent line, so you just need to find the slope. Begin by
finding the derivative of x
2
+ xy + y
2
= 3:
2
1
2
0
2
2
0
x
y x
dy
dx
y
dy
dx
x y x
dy
dx
y
dy
dx
+
+
+
=
+ +
+
=






Next, enter the values x = 1 and y = 1 and solve for the slope
dy
dx
:
2 1 1 1
2 1
0
3 3
0
1
( )
( )
( )
+ +
+
=
+
=
−
dy
dx
dy
dx
dy
dx
dy
dx
=
The tangent line has a slope of –1 and passes through (1, 1), so its equation is
y
x
y
x
− = −
−
= − +
1
1
1
2
(
)