Question

help please :( im confused..

Answer 1

oh, i did post it before.. but here it is again

Answer 2

wht class is this?

Answer 3

the slope of tangent line is dy/dx

Find dy/dx by using implicit differentiation

First i would simplify the equation by expanding the Left side ( the cubed terms should cancel)
\[x^3 + 3x^2 y + 3xy^2 + y^3 = x^3 + y^3\]
\[3x^2 y + 3xy^2 = 0\]
Not take derivative using product rule
\[\rightarrow 6xy + 3x^2 \frac{dy}{dx} + 3y^2 + 6xy \frac{dy}{dx} = 0\]
Group the dy/dx terms on Left side
\[3x^2 y \frac{dy}{dx} + 6xy \frac{dy}{dx} = -3y^2 -6xy\]
\[\frac{dy}{dx} = \frac{-3y^2 -6xy}{3x^2 y + 6xy} = \frac{-y -2x}{x^2 + 2x}\]
Next plug in the point where we want the slope of tangent line ..... x = -1 , y =?
\[\frac{dy}{dx} = y-2\]
Find y value using original equation
\[\rightarrow 3y -3y^2 = 0\]
\[\rightarrow y = 0,1\]
Therefore the 2 slopes are.
\[\frac{dy}{dx} = -1,-2\]

Answer 4

@stephy321: Next time, please post what you yourself have done towards finding a solution. That'd make it easier for others to step in and help you from the point at which you'd let off.