If $x ^ { 3 } - 2 x y + 3 y ^ { 2 } = 7$, then $\frac { d y } { d x } =$
(A) $\frac { 3 x ^ { 2 } + 4 y } { 2 x }$
(B) $\frac { 3 x ^ { 2 } - 2 y } { 2 x - 6 y }$
(C) $\frac { 3 x ^ { 2 } } { 2 x - 6 y }$
(D) $\frac { 3 x ^ { 2 } } { 2 - 6 y }$

On the coordinate plane, what is the slope of the tangent line to the curve $y ^ { 3 } = \ln \left( 5 - x ^ { 2 } \right) + x y + 4$ at the point $( 2,2 )$? [3 points]
(1) $- \frac { 3 } { 5 }$
(2) $- \frac { 1 } { 2 }$
(3) $- \frac { 2 } { 5 }$
(4) $- \frac { 3 } { 10 }$
(5) $- \frac { 1 } { 5 }$

Find the slope of the tangent line to the curve $2 x + x ^ { 2 } y - y ^ { 3 } = 2$ at the point $( 1,1 )$. [3 points]

What is the slope of the tangent line to the curve $e ^ { x } - x e ^ { y } = y$ at the point $( 0,1 )$? [3 points]
(1) $3 - e$
(2) $2 - e$
(3) $1 - e$
(4) $- e$
(5) $- 1 - e$

What is the slope of the tangent line to the curve $x ^ { 2 } - 3 x y + y ^ { 2 } = x$ at the point $( 1,0 )$? [3 points]
(1) $\frac { 1 } { 12 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 3 }$
(5) $\frac { 5 } { 12 }$

The normal to the curve $x^2 + 2xy - 3y^2 = 0$ at $(1, 1)$:
(1) does not meet the curve again
(2) meets the curve again in the second quadrant
(3) meets the curve again in the third quadrant
(4) meets the curve again in the fourth quadrant

The normal to the curve $x ^ { 2 } + 2 x y - 3 y ^ { 2 } = 0$, at $( 1,1 )$
(1) Meets the curve again in the fourth quadrant
(2) Does not meet the curve again
(3) Meets the curve again in the second quadrant
(4) Meets the curve again in the third quadrant

Let $y = y ( x )$ be a function of $x$ satisfying $y \sqrt { 1 - x ^ { 2 } } = k - x \sqrt { 1 - y ^ { 2 } }$ where $k$ is a constant and $y \left( \frac { 1 } { 2 } \right) = - \frac { 1 } { 4 }$. Then $\frac { d y } { d x }$ at $x = \frac { 1 } { 2 }$, is equal to
(1) $- \frac { \sqrt { 5 } } { 4 }$
(2) $- \frac { \sqrt { 5 } } { 2 }$
(3) $\frac { 2 } { \sqrt { 5 } }$
(4) $\frac { \sqrt { 5 } } { 2 }$

Which of the following points lies on the tangent to the curve $x^4 e^y + 2\sqrt{y+1} = 3$ at the point $(1, 0)$?
(1) $(2, 2)$
(2) $(2, 6)$
(3) $(-2, 6)$
(4) $(-2, 4)$

In the rectangular coordinate plane
$$y ^ { 2 } + \sin \left( x ^ { 2 } - 1 \right) = 4$$
What is the slope of the tangent line to the curve given by this equation at the point $\mathbf { P } ( - \mathbf { 1 } , - \mathbf { 2 } )$?
A) - 1
B) $\frac { 1 } { 2 }$
C) 2
D) $\frac { - 1 } { 2 }$