Is it correct to write that if $dP = \frac{-h\cdot d\lambda}{\lambda^2}$ then $\Delta P = \frac{-h\cdot \Delta \lambda}{\lambda^2}$?

Question

I was studying high school-level physics and came across a problem,

Which of the following relationships of Heisenberg's Uncertainty Principle are correct?

1. $\Delta P \cdot \Delta x \geq \frac{h}{4\pi}$
2. $\Delta E \cdot \Delta t \geq \frac{h}{4\pi}$
3. $\Delta \lambda\cdot\Delta x\leq \frac{-\lambda^2}{4\pi}$
4. $\Delta V \cdot \Delta x \geq \frac{h}{4\pi m}$

I was able to figure out that all are correct but for the third one, I derived it this way:

The momentum of a photon is $P=\frac{h}{\lambda}$, so if we differentiate it w.r.t. wavelength $\lambda$, we get $$\frac{dP}{d\lambda} = \frac{-h}{\lambda^2}$$$$\implies dP = \frac{-h\cdot d\lambda}{\lambda^2}$$ After that what I did is that I wrote, $$\Delta P = \frac{-h\cdot\Delta\lambda}{\lambda^2}$$ And substituted this in expression 1. Though, I don't think this is the correct way to do it. Can anyone tell me if it is correct? If not, how should we solve this? Thanks in advance!

Answer 1

Having $P = h/\lambda$ , we can get :
$P+\Delta P = h/(\lambda+\Delta\lambda)$

Hence :
$P+\Delta P - P = h/(\lambda+\Delta\lambda) - h/\lambda$

$\Delta P = h[1/(\lambda+\Delta\lambda) - 1/\lambda]$

$\Delta P = h[(\lambda - (\lambda+\Delta\lambda))/(\lambda+\Delta\lambda)\lambda]$

$\Delta P = h[ -\Delta\lambda/(\lambda^2+\lambda\Delta\lambda)]$

$(\lambda^2+\lambda\Delta\lambda) \approx (\lambda^2)$

Hence : $\Delta P = h[ -\Delta\lambda/\lambda^2]$

$\Delta P = -h\Delta\lambda/\lambda^2$