Find the slope of the tangent line to the curve $x ^ { 3 } + y ^ { 3 } = 6 x y$ at the point (3, 3).
A) $- 3$
B) $- 2$
C) $- 1$
D)0
E)1
F)2
G)3
H)4

Question

Find the slope of the tangent line to the curve $x ^ { 3 } + y ^ { 3 } = 6 x y$ at the point (3, 3).
A) $- 3$
B) $- 2$
C) $- 1$ 
D)0
E)1
F)2
G)3
H)4

Quizplus · Accepted Answer

To find the slope of the tangent line to the curve at a given point, we first need to implicitly differentiate the given equation with respect to $x$. The equation is $x^3 + y^3 = 6xy$. Differentiating both sides with respect to $x$, we get:$$3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$$Rearranging to solve for $\frac{dy}{dx}$, the derivative of $y$ with respect to $x$, gives:$$3y^2 \frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$$Factoring out $\frac{dy}{dx}$, we have:$$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$So,$$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$$Substituting $x = 3$ and $y = 3$ into the equation:$$\frac{dy}{dx} = \frac{6(3) - 3(3)^2}{3(3)^2 - 6(3)} = \frac{18 - 27}{27 - 18} = \frac{-9}{9} = -1$$Therefore, the slope of the tangent line at the point (3, 3) is $-1$.

Find the Slope of the Tangent Line to the Curve x3+y3=6xyx ^ { 3 } + y ^ { 3 } = 6 x yx3+y3=6xy

Find the Slope of the Tangent Line to the Curve $x ^ { 3 } + y ^ { 3 } = 6 x y$