Question

Use implicit differentiation to find an equation of the tangent line to the curve at the indicated point.

y = sin 3xy; (π/6,1)

Answer 1

`\bf y=sin(3xy)\qquad \left((\pi )/(6),1 \right)\\\\ -------------------------------\\\\ \cfrac{dy}{dx}=cos(3xy)\left[3\left[ y+x(dy)/(dx) \right] \right]\implies \cfrac{dy}{dx}=cos(3xy)\left[3y+3x(dy)/(dx) \right] \\\\\\ \cfrac{dy}{dx}=3ycos(3xy)+3xcos(3xy)\cfrac{dy}{dx} \\\\\\ \cfrac{dy}{dx}-3xcos(3xy)\cfrac{dy}{dx}=3ycos(3xy)\impliedby \textit{common factor}`


`\bf \cfrac{dy}{dx}[1-3xcos(3xy)]=3ycos(3xy)\implies \boxed{\cfrac{dy}{dx}=\cfrac{3ycos(3xy)}{1-3xcos(3xy)}} \\\\\\ \left. \cfrac{3ycos(3xy)}{1-3xcos(3xy)} \right|_{(\pi )/(6),1}\implies \cfrac{3\cdot 1\cdot cos\left( 3\cdot (\pi )/(6)\cdot 1 \right)}{1-3\cdot (\pi )/(6)cos\left(3\cdot (\pi )/(6)\cdot 1 \right)}\\\\\\ \cfrac{3\cdot 1\cdot cos\left( (\pi )/(2) \right)}{1-(\pi )/(2)cos\left( (\pi )/(2) \right)}=0\\\\ -------------------------------\\\\`


`\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-1=0(x-(\pi )/(6))\implies y-1=0 \\ \left. \qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ \boxed{y=1}`