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

Asked by jonathan valle on 30 Nov 2012
``` 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?

## Products

Answer by Azzi Abdelmalek on 30 Nov 2012
Edited by Azzi Abdelmalek on 2 Dec 2012

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
```

Azzi Abdelmalek on 2 Dec 2012
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
Jan Simon on 2 Dec 2012

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.

```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]
```
Azzi Abdelmalek on 2 Dec 2012

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 by Image Analyst on 30 Nov 2012