What happens to units when you take second derivative?
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Consider the basic definition of a derivative according to Leibniz:
ddxf(x)∣∣x=x0=limx→x0f(x)−f(x0)x−x0ddxf(x)|x=x0=limx→x0f(x)−f(x0)x−x0
From this you see that if f(x)f(x) has unit uu, and xx has unit vv, the derivative has unit uvuv.
More generally, the nn-th derivative dndxnf(x)dndxnf(x) has unit uvnuvn (this is easily shown by repeating the above argument).
EDIT: In addition to the mathematics consider a basic example: let a function s(t)s(t) denote the distance travelled by a car versus time, i.e. s(t)s(t) has unit meters mm and the time tt has unit seconds ss. The first derivative then has unit m/sm/s and descibes the velocity=distance per time. The second derivative has unit m/s2m/s2 and describes acceleration, which is [change in velocity] per time.
Translated to your graph, the first derivative corresponds to [change in gap probability per height] or [slope of gap probability]. The sought second derivate then denotes [change in [slope of gap probability] per height]
ddxf(x)∣∣x=x0=limx→x0f(x)−f(x0)x−x0ddxf(x)|x=x0=limx→x0f(x)−f(x0)x−x0
From this you see that if f(x)f(x) has unit uu, and xx has unit vv, the derivative has unit uvuv.
More generally, the nn-th derivative dndxnf(x)dndxnf(x) has unit uvnuvn (this is easily shown by repeating the above argument).
EDIT: In addition to the mathematics consider a basic example: let a function s(t)s(t) denote the distance travelled by a car versus time, i.e. s(t)s(t) has unit meters mm and the time tt has unit seconds ss. The first derivative then has unit m/sm/s and descibes the velocity=distance per time. The second derivative has unit m/s2m/s2 and describes acceleration, which is [change in velocity] per time.
Translated to your graph, the first derivative corresponds to [change in gap probability per height] or [slope of gap probability]. The sought second derivate then denotes [change in [slope of gap probability] per height]
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