If #y^2 - 2xy = 21#, what is the slope of the tangent at #(2, -3)#?

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Answer 1 · 2017-01-21T23:23:17

#dy/dx = 3/5# at #(2, -3)#.

Differentiate using implicit differentiation and the product rule. Remember that we are differentiating with respect to #x#.

#2y(dy/dx) - 2(y + x(dy/dx)) = 0#

#2y(dy/dx) - 2y - 2x(dy/dx) = 0#

#2y(dy/dx) - 2x(dy/dx) =2y#

#dy/dx(2y - 2x) = 2y#

#dy/dx = (2y)/(2(y - x))#

#dy/dx = y/(y - x)#

Simply evaluate now:

#dy/dx = -3/(-3 - 2)#

#dy/dx = -3/-5#

#dy/dx= 3/5#

Hopefully this helps!