# Implicit differentiation: two answers resulted

benhou
Could anyone explain that I got two different answers for this question: find $$dy/dx$$ of $$\frac{x}{x+y}-\frac{y}{x}=4$$.

1. using quotient rule:
$$\frac{x+y-(1+dy/dx)x}{(x+y)^{2}}-\frac{x\frac{dy}{dx}-y}{x^{2}}=0$$
$$\frac{y}{(x+y)^{2}}-\frac{x}{(x+y)^{2}}\frac{dy}{dx}+\frac{y}{x^{2}}-\frac{1}{x}\frac{dy}{dx}=0$$
$$(\frac{x}{(x+y)^{2}}+\frac{1}{x})\frac{dy}{dx}=\frac{y}{(x+y)^{2}}+\frac{y}{x^{2}}$$
$$x(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})\frac{dy}{dx}=y(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})$$
$$\frac{dy}{dx}=y/x$$

2. simplify using common denominator before taking derivative:
$$\frac{x^{2}}{x^{2}+xy}-\frac{xy+y^{2}}{x^{2}+xy}=4$$
$$x^{2}-xy-y^{2}=4x^{2}+4xy$$
$$-y^{2}=3x^{2}+5xy$$
$$-2y\frac{dy}{dx}=6x+5y+5x\frac{dy}{dx}$$
$$\frac{dy}{dx}=-\frac{6x+5y}{2y+5x}$$

Last edited:

Homework Helper
$$x^{2}-xy-y^{2}=4x^{2}+4xy$$
$$y^{2}=3x^{2}+3xy$$

The second line does not follow from the first.

benhou
Sorry, it's suppose to be "5xy"

benhou
and it's suppose to be -y^2. Now I corrected it.

benhou
Help, anyone? i still don't get why.