$\text { Find } \frac { d ^ { 2 } y } { d ^ { 2 } x } \text { in terms of } x \text { and } y \text { given that } x ^ { 2 } + 6 y ^ { 2 } = 9 \text {. Use the original equation to }$ simplify your answer.
A)$y ^ { tt } = - \frac { 1 } { 4 y ^ { 3 } }$
B) $y ^ { tt } = - 1 y ^ { 3 }$
C) $y ^ { tt } = - 4 y ^ { 3 }$
D) $y ^ { tt } = - 24 y ^ { 3 }$
E) $y ^ {tt } = - \frac { 1 } { 24 y ^ { 3 } }$

Question

Quizplus · Accepted Answer

Differentiating $x^2 + 6y^2 = 9$ with respect to $x$ gives $2x + 12yy' = 0$, leading to $y' = -\frac{x}{6y}$. Differentiating again with respect to $x$ gives $y'' = -\frac{6y(1) - (-x)(6y')}{(6y)^2} = -\frac{6y + x(-\frac{x}{6y})}{36y^2} = -\frac{6y^2 + x^2}{36y^3}$. Using the original equation $x^2 + 6y^2 = 9$, we substitute to get $y'' = -\frac{9}{36y^3} = -\frac{1}{4y^3}$.

Simplify Your Answer ytt=−14y3y ^ { tt } = - \frac { 1 } { 4 y ^ { 3 } }ytt=−4y31​

Simplify Your Answer $y ^ { tt } = - \frac { 1 } { 4 y ^ { 3 } }$