When calculating Percentage Uncertainty, do we need to make our interested variable as subject? Given that the quantities $L,x$ and $y$ are related by the equation $Lx=y^2$. What is the percentage uncertainty in $L$ if the percentage uncertainties in $x$ and $y$ are $1$% and $3$% respectively?
I process in two ways. 
First way, i straight away calculate $\frac{\Delta L}{L}$ without manipulating the equation $Lx=y^2$. In other words, I obtain 
$$\frac{\Delta L}{L} + \frac{\Delta x}{x} = 2\frac{\Delta y}{y}$$
Then I substitute values accordingly. I obtain $\frac{\Delta L}{L} = 5$%, which is not correct. 
Second way, I manipulate the equation into $L = \frac{y^2}{x}$. Then I apply the same reasoning, I obtain 
$$\frac{\Delta L}{L} = 7\%$$ 
This is the correct answer.
My question: Why would I get two different answers?
 A: There must be a mistake in your work, because both methods are equivalent and produce the same result, implying the equation $\frac{\Delta L}{L} = 2\frac{\Delta y}{y}-\frac{\Delta x}{x}$.
The issue is in following the rules too closely. If $L=y-x$, then $\Delta L=\Delta y-\Delta x$. This can't be put in a nice fractional form, but that doesn't matter. Let's say $y=1\pm 1$ and $x=0\pm 1$. If we took the second equation at face value, $\Delta L=0$, because $\Delta y=1$ and $\Delta x=1$. Clearly what we must do is consider the pair $-1\le \Delta y\le1$ and $-1\le \Delta x\le1$ that give the largest predicted value of $\Delta L$. In this case, that would be taking $\Delta y=1$ and $\Delta x=-1$.
As an example, take $x=7$ and $y=5$. Then $L=y^2/x=3.57$. Let's add the errors in. The first way gives a five percent error:
$$L=\frac{((1+0.03)\cdot 5)^2}{(1+0.01) \cdot 7}=3.751\approx (1+0.05)\times 3.57$$
But x can vary in the other direction and give a seven percent error:
$$L=\frac{((1+0.03)\cdot 5)^2}{(1-0.01) \cdot 7}=3.827\approx (1+0.07)\times 3.57$$
A: $Lx=y^2\Rightarrow \log L+\log x=2\log y$. Differentiating we get:
$\frac{dL}{L}+\frac{dx}{x}=2\frac{dy}{y}$. If the changes are small then approximately $\frac{\delta L}{L}+\frac{\delta x}{x}\approx2\frac{\delta y}{y}$. So the correct answer is 5%.
