Second partial derivatives of z² + sin x = tan y

[schattenjaeger]

I'm supposed to find (assume all these d's are the partial derivative sign, not d)

d^2z/dxdy, d^2z/dx^2, and d^2z/dy^2

The one I can't do is z^2 + sinx = tany

I set it equal to zero, so z^2 + sinx - tany=0

so dz/dx = - Fx/Fz = sec^2y/2z
dz/dy = - Fy/Fz = -cosx/2z

multiply them for d^2z/dxdy I get -(sec^2y*cosx)/(4z^2) which is ALMOST right, but the book says the denominator is 4z^3, and the other two I do the same procedure(dz/dx * dz/dx for the second one for example) and get nothing close

 

[mikeu]

dz/dx = - Fx/Fz = sec^2y/2z
dz/dy = - Fy/Fz = -cosx/2z

multiply them for d^2z/dxdy I get -(sec^2y*cosx)/(4z^2) which is ALMOST right, but the book says the denominator is 4z^3, and the other two I do the same procedure(dz/dx * dz/dx for the second one for example) and get nothing close

Your problem seems to be the fact that
$$\frac{\partial^2z}{\partial x\partial y} \neq \frac{\partial z}{\partial x}\frac{\partial z}{\partial y}.$$

On the left-hand side (what you want to find), you are taking the partial derivative with respect to x of the partial derivative of z with respect to y, that is
$$\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right).$$
This gives you a second-order derivative, whereas on the right-hand side (what you did) you are merely multiplying two first-order derivatives together... The same will be true for the other derivatives you have to find, since
$$\frac{\partial^2z}{\partial x^2} \neq \left(\frac{\partial z}{\partial x}\right)^2.$$

 

[schattenjaeger]

Ah, ok, one mistake outta the way

in that case, and I know I'm doing it wrong, if I take the partial derivative of z with respect to y, I end up with 2z*dz/dy + sec^2y, and if I take the partial derivative with respect to x at this point, I end up with 0, which clearly isn't right. I can usually do implicit differentiation but I'm doing something silly here

Edit, oh wait I'm stupid, I guess the -Fx/Fy stuff was still right then? Hang on

Edit #2 - Oh jeez, I had that bacwards up there too, so I'm still taking the partial with respect to x, but with no x in the equation which gives me 0:(

 

[schattenjaeger]

YES, got it, though I'm not sure it was the most efficient way
I set the equation equal to z to find dz/dy,
z=sqrt(tanx-sinx)
dz/dy=-((tany-sinx)^-1/2 / 2) * sec^2y

then derivative in terms of x, (-sec^2 y /2) * 1/2(tany-sinx)^-3/2 * cosx

plug in z for (tanx-sinx)^1/2 and that all simplifies to the right answer