Question

A) Find y' if x^3 + y^3 = 6xy.

Answer 1

A)

$\bf x^3+y^3=6xy\\\\ -----------------------------\\\\ 3x^2+3y^2\frac{dy}{dx}=6\left( 1\cdot y+x\frac{dy}{dx} \right)\\\\\\ 3\left( x^2+ y^2\frac{dy}{dx}\right)=6\left( 1\cdot y+x\frac{dy}{dx} \right) \\\\\\ x^2+ y^2\frac{dy}{dx}=2y+2x\frac{dy}{dx}\implies x^2+ y^2\frac{dy}{dx}-2y-2x\frac{dy}{dx}=0 \\\\\\ \cfrac{dy}{dx}(y^2-2x)=2y-x^2\implies \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x}$

B)

$\bf \left. \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x} \right|_{3,3}\implies \begin{cases} x=3\\ y=3 \end{cases}\implies -1 \\\\ -----------------------------\\\\ y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-3=-1(x-3)\\ \qquad \uparrow\\ \textit{point-slope form}$

solve for "y", and that'd be the equation of the tangent at 3,3

C)   recall, the tangent is horizontal when the slope is 0, thus

$\bf \cfrac{dy}{dx}=\cfrac{2y-x^2}{y^2-2x}\implies 0=\cfrac{2y-x^2}{y^2-2x}\implies 0=2y-x^2\implies \cfrac{x^2}{2}=\boxed{y} \\\\\\ now \qquad x^3+y^3=6xy\implies x^3+\left( \boxed{\cfrac{x^2}{2}} \right)^3=6x\left( \boxed{\cfrac{x^2}{2}} \right) \\\\\\ x^3+\cfrac{x^6}{8}=3x^3\implies \cfrac{8x^3+x^6}{8}=3x^3 \\\\\\ 8x^3+x^6-24x^3=0\implies x^6-16x^3=0 \\\\\\ x^3(x^2-16)=0\implies x=\{0,\pm 4\}$

so.. x =0 is not exactly in the first quadrant,  x = -4 isn't either

so. the only choice is really x = 4... what's "y" when x = 4? well, just plug that into the $\bf x^3+y^3=6xy\qquad or \qquad \cfrac{x^2}{2}=y$

and solve for "y"