1. Partial derivative

Having a bit of a problem with this partial derivative, i know how to do the problem if the x and y wasn't infront as you could apply the product rule and apply the partial derivative, however having problems in what to do with it, here is the problem:

f(x,y)=xcosx*coshy+ysinx*sinhy

Determine partial df/dx, df/dy, second partial df/dx, second partial df/dy and partial d^2f/dxdy

Many thanks in advance.

2. This problem may have arisen from a course in DE's, but it's really a calculus question. I'll do the first-order derivatives - see if you can finish with all the second-order derivatives. We have

$\displaystyle f(x,y)=x\cos(x)\cosh(y)+y\sin(x)\sinh(y).$ Then,

$\displaystyle \dfrac{\partial f}{\partial x}=(1\cdot \cos(x)-x\sin(x))\cosh(y)+y\cos(x)\sinh(y),$ and

$\displaystyle \dfrac{\partial f}{\partial y}=x\cos(x)\sinh(y)+\sin(x)(1\cdot\sinh(y)+y\cosh( y)).$

So you can see that I just used the usual product rule from Calc I to differentiate the products.

3. Originally Posted by Ackbeet
This problem may have arisen from a course in DE's, but it's really a calculus question. I'll do the first-order derivatives - see if you can finish with all the second-order derivatives. We have

$\displaystyle f(x,y)=x\cos(x)\cosh(y)+y\sin(x)\sinh(y).$ Then,

$\displaystyle \dfrac{\partial f}{\partial x}=(1\cdot \cos(x)-x\sin(x))\cosh(y)+y\cos(x)\sinh(y),$ and

$\displaystyle \dfrac{\partial f}{\partial y}=x\cos(x)\sinh(y)+\sin(x)(1\cdot\sinh(y)+y\cosh( y)).$

So you can see that I just used the usual product rule from Calc I to differentiate the products.
Fantastic i can see exactly where i went wrong, thanks very much. Just going out and will work on the second derivative later, thanks once more.

4. You're welcome. Let me know if you have any more difficulties.

5. Originally Posted by Ackbeet
You're welcome. Let me know if you have any more difficulties.
Applied it to other questions and getting them spot on now, however there is a second part where it says:
Without further explicit differentiation, determine partial (d^4f/dx^4)-(d^4f/dy^4) ive calculated the second partial derivative etc just wondering if there is any trick to it as it states without any further explicit differentiation.
Any help will be most appreciated.

6. Is this the same f as in the OP?

7. If you've calculated $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ you'll notice that

$\displaystyle f_{xx} + f_{yy} = 0$.

So how does that help with

$\displaystyle f_{xxxx} - f_{yyyy} = 0$?

Spoiler:

$\displaystyle f_{xxxx} - f_{yyyy} = 0 \;\text{can be written as}\; \left(f_{xx} + f_{yy}\right)_{xx} - \left(f_{xx} + f_{yy}\right)_{yy} = 0$ so your $\displaystyle f$ solves the $\displaystyle 4$th order PDE

8. Originally Posted by Danny
If you've calculated $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ you'll notice that

$\displaystyle f_{xx} + f_{yy} = 0$.

So how does that help with

$\displaystyle f_{xxxx} - f_{yyyy} = 0$?

Spoiler:

$\displaystyle f_{xxxx} - f_{yyyy} = 0 \;\text{can be written as}\; \left(f_{xx} + f_{yy}\right)_{xx} - \left(f_{xx} + f_{yy}\right)_{yy} = 0$ so your $\displaystyle f$ solves the $\displaystyle 4$th order PDE
I worked out partal d^2f/dx^2 and partial d^2f/dy^2 added together to be 0. Then wasn't sure where to go from there.
Many thanks in advance.