AP Calculus AB Quiz
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ANALYTICAL APPLICATIONS OF DIFFERENTIATION
Question 1
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For the relation
x
2
y
2
+
x
=
1
x^2y^2+x=1
x
2
y
2
+
x
=
1
, what is
d
y
d
x
\dfrac{dy}{dx}
d
x
d
y
in terms of
x
x
x
and
y
y
y
?
A
d
y
d
x
=
−
2
x
y
2
+
1
2
x
2
y
\dfrac{dy}{dx}=-\dfrac{2xy^2+1}{2x^2y}
d
x
d
y
=
−
2
x
2
y
2
x
y
2
+
1
B
d
y
d
x
=
−
2
x
y
2
2
x
2
y
\dfrac{dy}{dx}=-\dfrac{2xy^2}{2x^2y}
d
x
d
y
=
−
2
x
2
y
2
x
y
2
C
d
y
d
x
=
−
2
x
y
2
+
1
2
x
2
\dfrac{dy}{dx}=-\dfrac{2xy^2+1}{2x^2}
d
x
d
y
=
−
2
x
2
2
x
y
2
+
1
D
d
y
d
x
=
2
x
y
2
+
1
2
x
2
y
\dfrac{dy}{dx}=\dfrac{2xy^2+1}{2x^2y}
d
x
d
y
=
2
x
2
y
2
x
y
2
+
1
E
d
y
d
x
=
−
2
x
2
y
+
1
2
x
y
2
\dfrac{dy}{dx}=-\dfrac{2x^2y+1}{2xy^2}
d
x
d
y
=
−
2
x
y
2
2
x
2
y
+
1
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