How do you determine dy/dx given #y=x^2+xy#?

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Answer 1 · 2016-10-21T02:36:48

Given a choice, I would solve for #y# first.

#y=x^2+xy#

Making the function explicit
#y-xy=x^2#

#y = x^2/(1-x)#

#dy/dx = (2x(1-x^2)-x^2(-1))/(1-x)^2#

# = (2x-x^2)/(1-x)^2#

Leaving the function implicit

#y=x^2+xy#

#d/dx(y)=d/dx(x^2+xy)#

#dy/dx = 2x +y + x dy/dx#

#dy/dx = (2x+y)/(1-x)#

Answer 2 · 2016-10-21T02:41:22

#dy/dx=(2x+y)/(1-x)#

You have to use Implicit Differentiation, which is a special case of the chain rule, and the product rule.

#dy/dx=2x+x(1)dy/dx+y(1)#

Simplify

#dy/dx=2x+xdy/dx+y#

Solve for #dy/dx#

#dy/dx-xdy/dx=2x+y#

Factor out #dy/dx#

#dy/dx(1-x)=2x+y#

Isolate #dy/dx#

#(dy/dxcancel(1-x))/(cancel(1-x))=(2x+y)/(1-x)#

Simplify

#dy/dx=(2x+y)/(1-x)#

Here is an example of using the implicit differentiation.