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

Question

What is the slope of the tangent line to the curve $x y ^ { 3 } + y - 1 = 0$ at the point (0, 1)?
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 need to differentiate the equation of the curve with respect to $x$, using implicit differentiation. The given curve is $xy^3 + y - 1 = 0$. Differentiating both sides with respect to $x$, we get:$$y^3 + x(3y^2)\frac{dy}{dx} + \frac{dy}{dx} = 0$$At the point (0, 1), substituting $x = 0$ and $y = 1$, the equation simplifies to:$$1^3 + 0(3*1^2)\frac{dy}{dx} + \frac{dy}{dx} = 0$$$$1 + \frac{dy}{dx} = 0$$Solving for $\frac{dy}{dx}$, we find:$$\frac{dy}{dx} = -1$$Therefore, the slope of the tangent line to the curve at the point (0, 1) is $-1$.

What Is the Slope of the Tangent Line to the Curve