How do you find the slope of the line tangent to #x+y^3=y# at (0,0), (0,1), (0,-1), (-6,2)?

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Answer 1 · 2017-02-21T16:32:43

# (0,0) \ \ \ \ \ => dy/dx=1 #
# (0,1) \ \ \ \ \ => dy/dx=-1/2 #
# (0,-1) => dy/dx=-1/2 #
# (-6,2) => dy/dx=-1/11 #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

We have:

# x + y^3 = y #

Differentiating wrt #x# gives:

# 1 + 3y^2dy/dx = dy/dx #
# :. dy/dx - 3y^2dy/dx = 1 #
# :. dy/dx(1 - 3y^2) = 1 #

# :. dy/dx = 1/(1 - 3y^2) #

So the slope of the tangent at any point is the value of #dy/dx# at that point. thus;

# (0,0) \ \ \ \ \ => dy/dx=1 #
# (0,1) \ \ \ \ \ => dy/dx=-1/2 #
# (0,-1) => dy/dx=-1/2 #
# (-6,2) => dy/dx=-1/11 #

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