is possible use some function to find derivatives of a vector?

 by example:
NO2=( 1.1 2.4 3.3 4.7 5.9 6.0)' that corresponding to depth: Z=(4.5 6.2 8.4 10.3 12.5 14.8)' 
I want find d(NO2)/dz and d^2(NO2)/dz^2

Exist some function that calculate this?

Answer 1

Edit

NO2=[1.1 2.4 3.3 4.7 5.9 6.0]
Z=[4.5 6.2 8.4 10.3 12.5 14.8]
d1=diff(NO2)./diff(Z)
d2=diff(NO2,2)./diff(Z(2:end)).^2

Look at edit

Usually "d2/dz2" means the 2nd derivative, not a squaring.

dy and dt are not the same, dt don't change while dy is changing.

Try this for x-abssisse

t=1:10
diff(t,2)=
0 0 0 0 0 0 0 0

1. diff(y)./diff(t) is an approximation of the first derivative g=dy/dt, In general diff(t) is a constant, then diff(y)./diff(t)=cst*diff(y) , with cst=unique(1/diff(t))
2. the second derivative f=d(dy/dt)/dt is approximated by diff(cst*diff(y))./diff(t)=cst*diff(cst*diff(y))=cst^2*diff(diff(y))
3. finally f=diff(diff(y))./diff(t).^2=diff(y,2)./diff(t).^2

As far as I can see, your approximation is based on the assumption, that Z is equidistant. This is neither the general case, nor does it match the question. Therefore I think, that this approximation in unnecessarily rough, especially if the 2nd derivative is wanted.

Your method, cropped edges:

d2 =  [-0.0826, 0.1385, -0.0413, -0.2079]

Suggest 2nd order method, one-sided differences at the edges:

d2 = [-0.0912, -0.0563, 0.0126, -0.0555, -0.1354, -0.1116]

No, even Z is not equidistant, there is no reason that diff(Z) will change at each point, we are not looking for the variation of Z, it's No2. if the approximation is bad, it's because the distance between Z's value is big. To improve the result, maybe we can interpolate.

Answer 2

How about the gradient() function?

Answer 3

Matlab's GRADIENT is accurate in the fist order only for not equidistant input. See FEX: DGradient and FEX: central_difference.