Question

Partial Derivatives: Can someone please check my work?
http://screencast.com/t/nrWdgIXmIzK

Answer 1

you can simplify the ∂^2/∂y^2, a tiny bit

Answer 2

By factoring out a cosh(x+y) ?

Answer 3

cosh(x+y)+cosh(x+y)=2cosh(x+y)

Answer 4

ahh, didn't even see that :) thanks!

Answer 5

\[\large\frac{\partial ^{2}}{\partial y^{2}} [xy\sinh(x+y)] = x[2\cosh(x+y) + y\sinh(x+y)]\]

Answer 6

that is correct

Answer 7

Multiply or add?

Answer 8

add,

Answer 9

but i think you made a little error in the ∂^2/∂xy,

the x should be out side the square brackets right?

Answer 10

Putting the x outside of the square brackets doesn't change anything....?

Answer 11

ahh i see, i missed a round bracket at the end.

Answer 12

I think this has all the errors fixed http://screencast.com/t/OPDcKKiL1mbg

Answer 13

ok so you have got the ∂^2∂xy right but you can simplify further

Answer 14

\[\large\sinh(x+y)+ycosh(x+y)+[x(\cosh(x+y)+ysinh(x+y))]\] How do I simplify this further? @UnkleRhaukus

Answer 15

\[\sinh(x+y)+y\cosh(x+y)+[x(\cosh(x+y)+y\sinh(x+y))]\]\[=(1+xy)\sinh(x+y)+(y+x)\cosh(x+y)\]

Answer 16

ahhh,, thank you so much @UnkleRhaukus, if I could give you another medal I would. :(