How do you find dy/ dx for #4 + xy = y^2#?

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Answer 1 · 2015-08-30T14:41:47

Use implicit differentiation with the product and power rules.

Because of the topic under which this was posted (and because solving for #y# then differentiating is tedious),

let's differentiate implicitly.

#4 + xy = y^2#

Remember that #y# is some unknown function of #x#. It may help to think of it as "some stuff in parentheses".

#d/dx(4 + xy) = d/dx(y^2)#

#d/dx(4) + underbrace(d/dx(xy))_("product and chain")# #= underbrace(d/dx(y^2))_("power and chain")#

#0 + [underbrace((1)y + x d/dx(y))] = underbrace(2y^1 d/dx(y))#

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

Solve algebraically for

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

Note
The product rule can be written in various orders. I've used #d/dx(uv) = u'v+uv'#