1. ## Challening Differentiatioin Problem

If :
z=x²+2y²

While:
y=r sinθ
x= r cosθ

Find :-

1- (dz/dx)y
2- (dz/dx)r
3- (dz/dx)θ

4- (dz/dy)x
5- (dz/dy)r
6- (dz/dy)θ

7- (dz/dθ)x
8- (dz/dθ)y
9- (dz/dθ)r

10- (dz/dr)θ
11- (dz/dr)x
12- (dz/dr)y

13- d²z/dxdy
14- d²z/dxdθ
15- d²z/dydθ

16- d²z/drdx
17- d²z/drdθ
18- d²z/dxdy

2. Originally Posted by Simo
If : z=x²+2y²

While:
y=r sinθ
x= r cosθ

Find :-

1- (dz/dx)y
2- (dz/dx)r
3- (dz/dx)θ

4- (dz/dy)x
5- (dz/dy)r
6- (dz/dy)θ

7- (dz/dθ)x
8- (dz/dθ)y
9- (dz/dθ)r

10- (dz/dr)θ
11- (dz/dr)x
12- (dz/dr)y

13- d²z/dxdy
14- d²z/dxdθ
15- d²z/dydθ

16- d²z/drdx
17- d²z/drdθ
18- d²z/dxdy
Where are you stuck? Please be complete. Thank you!

3. Hello Simo:

Cheers

4. I presume that the letter after the derivative indicates that the derivative is to be taken holding that variable constant.

So here is the idea for the first few:

1) $\left(\frac{\partial z}{\partial x}\right)_y$

Since $z= x^2+ 2y^2$. Holding y constant, the derivative with respect to x is 2x.

2) $\left(\frac{\partial z}{\partial x}\right)_r$
$x^2+ y^2= r^2cos^2(\theta)+ r^2 sin^2(\theta)= r^2(cos^2(\theta)+ sin^2(\theta))= r^2$ so $y^2= r^2- x^2$
$x^2+ 2y^2= x^2+ 2(r^2- x^2)= -x^2+ 2r^2$. Holding r constant, the derivative with respect to x is -2x.

3) $\left(\frac{\partial z}{\partial x}\right)_\theta$
$\frac{y}{x}= \frac{r sin(\theta)}{r cos(\theta)}= tan(\theta)$ so $y= x tan(\theta)$ and $x^2+ 2y^2= x^2+ 2x^2 tan^2(\theta)= x^2(1+ 2 tan^2(\theta))$. Holding $\theta$ constant the derivative with respect to x is $2x(1+ 2tan^2(\theta))$

Now, YOU try the others!