How do you find the derivative of #x+xy=y^2#?

1 Answer
Jim H
May 3, 2016

I assume that you want the derivative of #y# with respect to #x#. I get #dy/dx = (1+y)/(2y-x)#

Explanation:

There is no such thing as the derivative of an equation. We can find the derivative of either variable with respect to a variable. I assume that you want to find #dy/dx#.

#x+xy=y^2#

We assume that #y# is some function of #x# that we haven't found. Think of it as "some stuff in parentheses".

#underbrace(x)_"term 1"+underbrace(x("some stuff"))_"term 2" = underbrace(("some stuff")^2)_"term 3"#

We have 3 terms and we will differentiate term-by-term.

In order to differentiate #x("some stuff")#, we'll need the product rule and for #("some stuff")^2# we'll need the power rule and the chain rule.

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

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

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

There are 4 terms now. Two of them include a factor of #dy/dx# and the other two do not. Solve for #dy/dx#.

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

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

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

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