Solve for dy/dx- multiple answers?

[Musclemanjr]

Hi everyone,

I was taking a calculus test today and I had some time left over, so I checked my work on a problem:

Solve for dy/dx:
y=x/(x+y)

Since the unit is over product/quotient rule, I naturally used one. I checked my answer with the other:

Quotient rule:

1) y=x(x+y)

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

3) dy/dx = (x + y - x - x*dy/dx)/((x + y)^2)

4) dy/dx = (y - x*dy/dx)(x + y)^-2

5)dy/dx*(x + y)^2 = y - x*dy/dx

6)dy/dx*(x + y)^2 + x(dy/dx) = y

7) dy/dx*((x + y)^2 + x) = y

8) dy/dx = y/(x + y)^2 + x)

I checked my work using the product rule:

1) y = x/(x + y)

2) y = x*(x + y)^-1

3) dy/dx = (1)(x + y)^-1 + x*(-1*((x + y)^-2)*(1 + dy/dx))

4) dy/dx = 1/(x + y) + -x*((1 + dy/dx)/((x + y)^2))

5) dy/dx = 1/(x + y) + (-x - x*dy/dx)/((x + y)^2)

6) dy/dx*(x + y)^2 = (x + y) + (-x - x*dy/dx)

7) dy/dx*(x + y)^2 = y - x*dy/dx

8) dy/dx*(x + y)^2 + x*dy/dx = y

9) dy/dx*((x+y)^2+x) = y

10) dy/dx = y/((x+y)^2+x)

So the product rule and the quotient rule gave me the same answer. However, I tried it another way by moving around the original equation before differentiating:

1) y = x/(x+y)

2) y*(x+y) = x

3) xy + y^2 = x

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

5) dy/dx*x + 2y*dy/dx = 1-y

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

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

Doing the problem this way yields a completely different answer. I was wondering if anybody could give me some insight as to why this is or the correct answer.

Thanks in advance.

 

[soroban]

Hello, Musclemanjr"!

I cranked out the three derivatives and got the same results.

\(\displaystyle \text{I'm still trying to prove that: }\:\frac{dy}{dx} \;=\;\frac{y}{(x+y)^2+x} \;=\;\frac{1-y}{x+2y}\)



Want some disturbing news?

I differentiate another way and came up with yet another answer!


\(\displaystyle \text{We have: }\;y \:=\:\frac{x}{x+y} \quad\Rightarrow\quad y(x+y) \:=\:x \quad\Rightarrow\quad xy+ y^2 \:=\:x\)

. . . . . . . \(\displaystyle x - xy \:=\:y^2 \quad\Rightarrow\quad x(1-y) \:=\:y^2 \quad\Rightarrow\quad x \:=\:\frac{y^2}{1-y}\)


\(\displaystyle \text{Differentiate with respect to }y\!:\;\;\frac{dx}{dy} \;=\;\frac{2y(1-y) - y^2(-1)}{(1-y)^2} \;=\;\frac{2y-y^2}{(1-y)^2}\)

. . \(\displaystyle \text{Therefore: }\;\frac{dy}{dx} \;=\;\frac{(1-y)^2}{y(2-y)}\)