
Urgent help please
How can i write a system between 2 derivates with the ti89?
I mean
d/dx [f(x,y)=a]
where
dy/dx=c
for example if f(x,y)=a is x^2+3 =y
then the solution is 2x=c => x=c/2
I wasn't able to find anything... i've tried to do something like
solve(d(x^2+3=y,x)(d(y,x)=1),x) >wrong value
d(x^2+3=y,x)(d(y,x)=1)) >2x=0 wrong
Thanks in advance
Marco

Hello,
Instead of using , separate your equations with and :
Code:
solve(d(x^2+3,x)=y and d(y,x)=1,x)
The problem is that they rather consider y as a variable than a function. This is partly why there is an error.
There are huge problems in your syntax. For systems of equations, you must have : solve(...=!!! and :::=^^^,{variable1,variable2})
After thinking about it, I don't really know...

that equation give me a "false", maybe i was a bit unclear (i'm not english / american)..
I have this equation:
x^2+3 =y >d/dx> 2x=dy/dx
now i impose dy/dx=c (it's the same thing i have a system of 2 equations)
then the equation become 2x=c => x=c/2
I think there should be a way to tell to the ti89 to leave d/dx when it find derivates on unknown quantity/costant...

I've find a way, but it's not right at all..
d(x^2+3=yx=cy) > 2x=c
but in this case i think calc. will calculate the system firstly, and after it will do the derivate... so this is wrong too

I've found the solution.
d(x^2+3=f(x),x) > 2x=df(x)/dx
then by hands i write df(x)/dx=c so x=c/2.
Well, in this way it seems completely useless, but with giant eq. i usually resolve, it's a pretty nice way.