To transform the variables "solve" and to do the back-substitution "subs"...



2013/11/2 F. B. <franz.bona...@gmail.com>

>
>
> On Saturday, November 2, 2013 11:46:27 AM UTC+1, Saullo Castro wrote:
>>
>> 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?
>>
>
> Yes, it is.
>
> Are you going to use .subs( ) or solve( ) ?
>
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