just adding... after `x2.diff(v)` the `v` must be replaced by `a + c**2`
again...



2013/11/2 Saullo Castro <[email protected]>

> I believe we can make a variable transform and then apply the derivative
> using the expression converted to a variable, like:
>
>     x = a**2 + c + d**3
>     x.diff((a + c**2))
>
> changing variables:
>
>     v = a + c**2
>     a = v - c**2
>     c = (v-a)**0.5
>
> the new x will be:
>
>     x2 = (v-c**2)**2 + (v-a)**0.5 + d**3
>
> and the derivative could be computed as:
>
>     x2.diff(v)
>
> is that reasonable?
>
>
> 2013/11/2 F. B. <[email protected]>
>
>
>>
>> On Saturday, November 2, 2013 12:23:00 AM UTC+1, brombo wrote:
>>>
>>>  Consider Lagrangian field theory where the derivatives are taken with
>>> respect to the gradient of a field.  In the case of quantum electrodynamics
>>> with respect to the gradient of a spinor field.
>>>
>>
>> Yes, that would be needed. I am wondering, is there a generic algorithm
>> for functional derivatives, or is it more likely to be a complicated matter?
>>
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