if z= cos(7x-3y)sin(2x+8y) show d^2z/dxdy & d^2z/dxdy.
Can anyone give me the solution for this as the one i came up with looks very obscure!!
Thanks in advance
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if z= cos(7x-3y)sin(2x+8y) show d^2z/dxdy & d^2z/dxdy.
Can anyone give me the solution for this as the one i came up with looks very obscure!!
Thanks in advance
It's too boring to write out. If you're still stuck I'll scan the solution and post it up on the www somewhere.
I do not think that order matters when doing partial differentation. I think d^2z/dydx = d^2/dxdy
BTW: You asked for the same derivative twice. I assumed a typo.
If you know how to find an ordinary derivative, you can work out partials.
For z = Function(x, y), partial dz/dx is determined by treating y as a constant. Similarly for dz/dy: Trreat x as a constant.
For d^2z/dxdy, find dz/dx as above and then do d(dz/dx)/dy treating x as a constant.
I, too, am too lazy to write it out.
Find what x equals in terms of z
Find what y equals in terms of z
Find dz/dx
Find dz/dy
multiply. That should work.
Ambivalentiowa: That algorithm is incorrect. You either made it up or you copied it incorrectly and from the wrong context.
It is not nice to post invalid data here. If you are not certain, check some reliable source.
Unlike the Chit Chat forum, some people here use the data posted for serious work and we should not mislead them. We should also try to avoid confusing them, but this is not always possible.
You must differentiate twice to get a second partial derivative.
If z = FunctionA(y) and y = FunctionB(x) then
dz/dx = (dz/dy)*(dy/dx)
I am not sure of the above. I know there is some situation where you can multiply derivatives, but it does not relate to partials.
I'm sorry Guv. That how i did it on my homework so i must have got that wrong. Anyway, i have been very helpful in other posts so i don't think one wrong post should get me a reprimand.