Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
$y \sin 3 x = x \cos 3 y , \left( \frac { \pi } { 3 } , \frac { \pi } { 6 } \right)$
A) $y = \frac { x } { 2 }$
B) $y = 2 x - \frac { 3 \pi } { 3 }$
C) $y = - \frac { x } { 2 } + \frac { \pi } { 2 }$
D) $y = \frac { x } { 6 }$
E) $y = \frac { x } { 3 } + \frac { \pi } { 6 }$

Question

Quizplus · Accepted Answer

Implicit differentiation gives$$y' \sin 3x + y \cos 3x (3) = x' \cos 3y - x \sin 3y (3)$$$$y' \sin 3x + 3y \cos 3x = \cos 3y - 3x \sin 3y$$$$y' = \frac{\cos 3y - 3x \sin 3y - 3y \cos 3x}{\sin 3x}$$At the point $(\frac{\pi}{3}, \frac{\pi}{6})$,$$y' = \frac{\cos \pi - 3\left(\frac{\pi}{3}\right) \sin \pi - 3\left(\frac{\pi}{6}\right) \cos \pi}{\sin \pi} = \frac{1 - \pi}{0} = \text{ undefined}$$So the tangent line is vertical and its equation is $x = \frac{\pi}{3}$.

Use Implicit Differentiation to Find an Equation of the Tangent ysin⁡3x=xcos⁡3y,(π3,π6)y \sin 3 x = x \cos 3 y , \left( \frac { \pi } { 3 } , \frac { \pi } { 6 } \right)ysin3x=xcos3y,(3π​,6π​)

Use Implicit Differentiation to Find an Equation of the Tangent $y \sin 3 x = x \cos 3 y , \left( \frac { \pi } { 3 } , \frac { \pi } { 6 } \right)$