How Do You Calculate the Uncertainty in the Derivative of y = a / (x - b)^3?

[Xizel]

y = a / (x - b)³

a = 77.1 ± 15.2
b = -1.78 ± 1.18
x = 21 ± 1

---

y' = -3a / (x - b)⁴

How do I find the uncertainty of y'? Thanks.

 

[Mark44]

y = a / (x - b)³

a = 77.1 ± 15.2
b = -1.78 ± 1.18
x = 21 ± 1

---

y' = -3a / (x - b)⁴

How do I find the uncertainty of y'? Thanks.

You can use the differential of y (dy) to approximate the uncertainty in y. Here, because a, b, and x are varying, what you have is y' = f(a, b, x) = $\frac {-3a} {(x - b)^4}$.
So $\Delta y' \approx dy' = \frac{\partial f}{\partial a} da + \frac{\partial f}{\partial b} db + \frac{\partial f}{\partial x} dx$
For da, db, dx, use the ± values above.

 

[Ray Vickson]

y = a / (x - b)³

a = 77.1 ± 15.2
b = -1.78 ± 1.18
x = 21 ± 1

---

y' = -3a / (x - b)⁴

How do I find the uncertainty of y'? Thanks.

You can compute y' for all 8 values a = 77.1 - 15.2, 77.1 + 15.2, b = -1.78 - 1.18, -1.78 + 1.18, x = 21-1, 21+1. Since that might be tedious, you can try to figure out ahead of time where the extreme values of y' will be found. The largest value of $|y'| = 3 a / |x-b|^4$ will occur when the numerator is as large as possible and the denominator is a small as possible. The minimum of $|y'|$ will be found from conditions opposite to those above.

I am a bit skeptical about using the formula for $\Delta y'$ given in #2, because in this case the $da, db$ are not "small" (but, perhaps, $dx$ is small enough).

 

[Mark44]

I am a bit skeptical about using the formula for $\Delta y'$ given in #2, because in this case the $da, db$ are not "small" (but, perhaps, $dx$ is small enough).

That thought also occurred to me. Even $\Delta x$ is not all that small, at about 5% of x. The increments for a and b are much worse.