# Redefine symbolic function even in derivatives

As a simple example, I have the following variables and functions:

```
r = var('r')
th = var('th', latex_name = '\\theta')
g = function('g')(r, th)
f = g + diff(g,r)
```

**From now on** I want to decompose `g`

into `g0`

and `g2`

as:

```
g0 = function('g0', latex_name = 'g_0')(r)
g2 = function('g2', latex_name = 'g_2')(r)
g = g0 + cos(th)*g2
```

If I print `f`

I still get:

$ f = g(r, \theta) + dg(r,\theta)/dr $

So instead, I apply a substitution on `g`

:

```
f = f.subs(g==g0+cos(th)*g2)
```

This changes `g`

, but not `diff(g,r)`

:

$f = g_0(r) + g_2(r)*\cos(\theta) + dg(r,\theta)/dr$

In order to change also `diff(g,r)`

I have to substitute explicitly the derivative:

```
f = f.subs(diff(g,r) == diff(g0+cos(th)*g2, r))
```

My question is the following:

Is there a way I can redefine a function without having to redefine also every single derivative? This way, I would avoid having to write all these substitutions:

```
diff(g,r) == diff(g0+g2*cos(th), r),
diff(g,r,r) == diff(g0+g2*cos(th), r, r),
diff(g,th) == diff(g0+g2*cos(th), th),
diff(g,th,th) == diff(g0+g2*cos(th), th, th),
diff(g,th,r) == diff(g0+g2*cos(th), th, r),
diff(g,r,th) == diff(g0+g2*cos(th), r, th)
```