$\text { Find } \frac { d y } { d x } \text { by implicit differentiation. }$ $x ^ { 3 } + 8 x + x ^ { 13 } y - y ^ { 6 } = 4$
A)$\frac { d y } { d x } = \frac { 3 x ^ { 2 } + 8 + 13 x ^ { 12 } y } { 6 y ^ { 5 } - x ^ { 13 } }$
B)$\frac { d y } { d x } = \frac { 3 x ^ { 2 } + 8 + 13 y } { 5 y ^ { 5 } - 13 x }$
C)$\frac { d y } { d x } = \frac { 3 x ^ { 2 } + 8 + 13 x ^ { 12 } y } { 5 y ^ { 5 } - x ^ { 13 } }$
D)$\frac { d y } { d x } = \frac { 3 x ^ { 2 } + 8 + x ^ { 13 } y } { 6 y ^ { 5 } - x ^ { 13 } }$
E)$\frac { d y } { d x } = \frac { 3 x ^ { 2 } - 8 + 13 x ^ { 12 } y } { 6 y ^ { 5 } - x ^ { 13 } }$

Question

Quizplus · Accepted Answer

Differentiating both sides of the equation $x^3 + 8x + x^{13}y - y^6 = 4$ with respect to $x$, using the product rule for $x^{13}y$ and the chain rule for $y^6$, gives $\frac{dy}{dx} = \frac{3x^2 + 8 + 13x^{12}y}{6y^5 - x^{13}}$.

Find dydx by implicit differentiation. \text { Find } \frac { d y } { d x } \text { by implicit differentiation. } Find dxdy​ by implicit differentiation.

$\text { Find } \frac { d y } { d x } \text { by implicit differentiation. }$