What is the slope of the tangent line of #(1-x)/(3-4y^2)-y = C #, where C is an arbitrary constant, at #(1,1)#?

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Answer 1 · 2016-02-03T14:51:01

The slope is #1#

This looks too tedious as written, so I'll rewrite.

Given that #(1,1)# lies on the curve, #C# is not arbitrary, but can be found:

#(1-(1))/(3-4(1)^2) - (1) = C#, so #C = -1#.

Therefore, if we replace #C# by #-1# and multiply through by #3-4y^2#, we get

#1-x-3y+4y^3=4y^2-3#.

Differentiating implicitly with respect to #x# gets us

#-1-3 dy/dx + 12y^2 dy/dx = 8y dy/dx#.

Now, find #dy/dx# at #(1,1)#

#-1 + (-3+12-8)dy/dx = 0#

So #dy/dx = 1#