Propagation of Errors - Error on Gradient

[FishareFriend]

Homework Statement

So I'm trying to calculate the error on the gradient I've obtained for my lab work. The line of best fit is too precise to use the parallelogram method and I'm still at the stage of my course where calculations of the gradient and such must be done by hand and not using a plotting program. So my question is, can I used the propagation of errors method for:
[itex]Y=a/b[/itex]
on the value used to calculate the gradient, i.e. the number I got from drawing a triangle under the line of best fit and taking one value from the other to get Δy and Δx, and those number's respective errors which I have from the various equipment I used to calculate the values.
Hence I'll have:
[itex]∂grad=grad*\sqrt{(∂x/Δx)^2+(∂y/Δy)^2}[/itex]

It's more just is this allowed as a way of calculating the error on the gradient or should I use a different method? If so which?

Homework Equations

Propagation of Errors for:
[itex]Y=a/b[/itex]

[itex]∂Y=Y*\sqrt{(∂a/a)^2+(∂b/b)^2}[/itex]

The Attempt at a Solution

 

[haruspex]

[itex]∂grad=grad*\sqrt{(∂x/Δx)^2+(∂y/Δy)^2}[/itex]

Not sure I understand your equation. If there are just two datapoints, I could interpret that as Δx and Δy being the horizontal and vertical displacements. Then grad = Δy/Δx, and δgrad = (Δy+δy)/(Δx+δx) - Δy/Δx ≈ (δy-grad.δx)/Δx = grad*(δy/Δy-δx/Δx). But I'm unclear what you mean by Δx and Δy in a more general case.
It sounds like you are taking Δx to be the full range of x values and setting Δy = grad*Δx.