Question

let z=z(x,y) be given imlicitely by sin(x+y)+cos(x+y)=pi/4 find Zxy..

Answer 1

I don't get what Zxy means

Answer 2

Can you tell me what it means please?

Answer 3

it is derivative of Z wrt. x first and second derivative of y

Answer 4

oh I think I understand
sorry

Answer 5

So we need Zx first!

Answer 6

\[(x+y)_xcos(x+y)+(x+y)_x(-\sin(x+y))=0\]

\[(1+0)\cos(x+y)+(1+0)(-\sin(x+y))=0\]

\[1\cos(x+y)+1(-\sin(x+y))=0\]

\[\cos(x+y)-\sin(x+y)=0\]


Since I was taking the partial derivative w.r.t. x I treated y as a constant
Do you see that?

Answer 7

So that is Zx

Answer 8

We need to do
(Zx)y now

Answer 9

So we will look at Zx
and we will take partial derivative of it with respect to y
that means we will treat x like it is a constant

Answer 10

Do you want to try?

Answer 11

-cos(x+y)-sin(x+y)=0 right ?

Answer 12

\[(x+y)_y(-\sin(x+y))-(x+y)_ycos(x+y)=0\]
\[(0+1)(-\sin(x+y)-(0+1)\cos(x+y)=0\]
\[-\sin(x+y)-\cos(x+y)=0\]
yes that is right
that is Zxy