#Question about Second Order Partial Derivatives
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for this question what second order derivatives would I need
i have u_x, u_y, u_z and then the second order u_xx, u_yy, u_zz so far
do I need to do all combinations for it
like u_xy, u_xz, u_zy?
So here, you can see it as a tree that you need to complete.
0-th order: u
1st order: u_x, u_y, u_z
2nd order: u_xx, u_xy, y_xz, u_yx, u_yy, u_yz, u_zx, u_zy, u_zz
ðŸ˜
Do note however that in the second order, many of them are identical
I forgot the name of the theorem
yeah thats what I was thinking
I went over that in class so I thought they'd be the same
So in practice, for the second order, you can reduce it from 9 to like, 6
since for instance u_xy = u_yx
it's a bit tedious, you can't avoid that
on the bright side the function is relatively simple to differentiate
So grit your teeth for 2 minutes and you'll be done
yeah the x and z ones differentiate the same just different letters
in my test last week I forgot that differentiating (ln(x))' = (x)'/x
so I just had it written as say (ln(5x))' = 1/5x
well, you're missing points on writing the equal sign
because ln(5x) is definitely not equal to 1/5x
and secondly, seeing that ln(5x) = ln(5) + ln(x), this is quite simple to differentiate

it was some double chain rule equation :(
with sin inside so it fried my brain a bit
yes, but calculus is better when you don't have to calculate

Anyway, do you still need help for this problem?
no i think I got it
Schwartz theorem
thank you guys
Dankeschön
i forgor how to rep
In that case, please close the post
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