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

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Answer 1 · 2018-05-03T16:33:41

#m = 1/10#

Given that the curve contains the point #(1,1)#, we find #C = 9#

The equation is equivalent to

#x^2+4xy+4y^2 = 9e^(x-y^2)#.

Differentiate implicitly to get

#2x+4y+4xdy/dx+8ydy/dx=9e^(x-y^2)(1-2ydy/dx)#

At #(1,1)# we have

#2+4+4dy/dx+8dy/dx=9-18dy/dx#

so, #dy/dx = 1/10#