benhou
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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}
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}
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