- #1

benhou

- 123

- 1

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

1. using quotient rule:

[tex]\frac{x+y-(1+dy/dx)x}{(x+y)^{2}}-\frac{x\frac{dy}{dx}-y}{x^{2}}=0[/tex]

[tex]\frac{y}{(x+y)^{2}}-\frac{x}{(x+y)^{2}}\frac{dy}{dx}+\frac{y}{x^{2}}-\frac{1}{x}\frac{dy}{dx}=0[/tex]

[tex](\frac{x}{(x+y)^{2}}+\frac{1}{x})\frac{dy}{dx}=\frac{y}{(x+y)^{2}}+\frac{y}{x^{2}}[/tex]

[tex]x(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})\frac{dy}{dx}=y(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})[/tex]

[tex]\frac{dy}{dx}=y/x[/tex]

2. simplify using common denominator before taking derivative:

[tex]\frac{x^{2}}{x^{2}+xy}-\frac{xy+y^{2}}{x^{2}+xy}=4[/tex]

[tex]x^{2}-xy-y^{2}=4x^{2}+4xy[/tex]

[tex]-y^{2}=3x^{2}+5xy[/tex]

[tex]-2y\frac{dy}{dx}=6x+5y+5x\frac{dy}{dx}[/tex]

[tex]\frac{dy}{dx}=-\frac{6x+5y}{2y+5x}[/tex]

1. using quotient rule:

[tex]\frac{x+y-(1+dy/dx)x}{(x+y)^{2}}-\frac{x\frac{dy}{dx}-y}{x^{2}}=0[/tex]

[tex]\frac{y}{(x+y)^{2}}-\frac{x}{(x+y)^{2}}\frac{dy}{dx}+\frac{y}{x^{2}}-\frac{1}{x}\frac{dy}{dx}=0[/tex]

[tex](\frac{x}{(x+y)^{2}}+\frac{1}{x})\frac{dy}{dx}=\frac{y}{(x+y)^{2}}+\frac{y}{x^{2}}[/tex]

[tex]x(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})\frac{dy}{dx}=y(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})[/tex]

[tex]\frac{dy}{dx}=y/x[/tex]

2. simplify using common denominator before taking derivative:

[tex]\frac{x^{2}}{x^{2}+xy}-\frac{xy+y^{2}}{x^{2}+xy}=4[/tex]

[tex]x^{2}-xy-y^{2}=4x^{2}+4xy[/tex]

[tex]-y^{2}=3x^{2}+5xy[/tex]

[tex]-2y\frac{dy}{dx}=6x+5y+5x\frac{dy}{dx}[/tex]

[tex]\frac{dy}{dx}=-\frac{6x+5y}{2y+5x}[/tex]

Last edited: