How do you use implicit differentiation to find dy/dx given #ln(x^2+y^2-xy)=100#?

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Answer 1 · 2016-11-13T15:17:09

#x^2 + y^2 - xy = e^100#

We use the product rule, power rule, implicit differentiation and the rule #(e^(f(x)))' = f'(x) xx e^(f(x))#

#2x + 2y(dy/dx) - (y + x(dy/dx)) = 0(e^100)#

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

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

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

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

Hopefully this helps!