Heisenberg's Uncertainty Principle - finding uncertainty in wavelength I am confused about this problem:

I needed to find the uncertainty of a wavelength using Heisenberg's uncertainty principle. In the solution, they differentiated $λ=c/f$ with respect to frequency to get $\frac{-c}{f^{2}}$.  λ is then substituted back in to get $\frac{- λ^{2}}{c}$. $Δf$ had already been found, so the final equation to find $Δλ$ was:
$$Δλ = \frac{- λ^{2}}{c}*Δf$$
Which honestly makes no sense to me. Why is it necessary to differentiate λ? I was thinking that the final equation would be something like:
$$Δλ = \frac{c}{Δf}$$
since $λ = \frac{c}{f}$. Why wouldn't my method work?
 A: As to why differentiation, we are talking about basic calculus.  You want to find out how $\lambda$ varies with $f$.
$\lambda =\frac{c}{f}$
$\frac{d\lambda}{df}=-\frac{c}{f^2}=-\frac{\lambda ^2}{c}$
$d\lambda=-\frac{\lambda ^2}{c}\ df$
Change the differentials to $\Delta$'s and you get what they have.  Your way is incorrect in that 
$d\lambda\neq \frac{c}{df}$
A: A wavelength has dimensions of a length: One way to apply the uncertainty principle is by application of an uncertainty between space and time:
$\Delta t \Delta x$
and so the uncertainty is a length squared quantity
$c \Delta t \Delta x \geq \frac{\hbar}{2}$
This is the spacetime uncertainty principle. Interpretation into the wavelength is using natural units now of $c = 1$
$\Delta t \Delta \lambda \geq \frac{\hbar}{2}$
edit: A poster has rightfully pointed out the dimensions are not entirely correct, but this was because I was attempting to teach the poster about dimensions. This edit is now to show the correct formula:
$\int q \cdot dp = action$
In generalized coordinates in which the wavelength (times) the momentum gives you an action. I fear people are quick to put down a post and not realize when someone is actually trying to help. Outside of generalized coordinates it can also be seen as
$\int \lambda \cdot dp = action$
or even
$\int p \cdot \Delta \lambda \geq \frac{\hbar}{2}$
Be aware though, that the order of multiplication between these quantities is always important. So why... simply because they do not commute
$[p,x] \geq \hbar$
