If $z=f(x,y$) where $x=e^u \cos(v)$ and $y=e^u \sin(v)$ then the value of $y\frac{∂z}{∂u}+x\frac{∂z}{∂v}$?

Question

If $z=f(x,y$) where $x=e^u \cos(v)$ and $y=e^u \sin(v)$ then the value of

$y\frac{∂z}{∂u}+x\frac{∂z}{∂v}$ is

$(A)e^{2u}\frac{∂z}{∂y}\quad(B) e^{2u}\frac{∂z}{∂x}\quad (C)e^{2u}(\frac{∂z}{∂x}+\frac{∂z}{∂y})\quad (D)0$

I tried it by applying the chain rule

$y\frac{∂z}{∂u}=y\frac{∂z}{∂x}\frac{∂x}{∂u} \tag{1}$

$x\frac{∂z}{∂v}=x\frac{∂z}{∂y}\frac{∂y}{∂u}. \tag{2}$

Adding $(1)$ and $(2)$ we get

$y\frac{∂z}{∂u}+x\frac{∂z}{∂v}=y\frac{∂z}{∂x}\frac{∂x}{∂u}+x\frac{∂z}{∂y}\frac{∂y}{∂u}\tag{3}.$

As $\frac{∂x}{∂u}=e^u \cos(v)$ & $\frac{∂y}{∂v}=e^u \cos(v)$, putting these values in $(3)$, we get

$y\frac{∂z}{∂u}+x\frac{∂z}{∂v}=e^u \cos(v)[y\frac{∂z}{∂x}+x\frac{∂z}{∂y}].$

My answer is not matching with any of the given options.

Where I'm committing mistake??

Answer 1

Since $z$ is a function of two independent variable $x$ and $y$, so $\frac{∂z}{∂u}=\frac{∂z}{∂x}\frac{∂x}{∂u}+\frac{∂z}{∂y}\frac{∂y}{∂u} \tag{1}$ and similarly $\frac{∂z}{∂v}=\frac{∂z}{∂x}\frac{∂x}{∂v}+\frac{∂z}{∂y}\frac{∂y}{∂v}. \tag{2}$ Here you are wrong. For more details see the pdf.

Therefore using $(1)$ and $(2)$, $yz_u+xz_v=e^uz_x(e^usinv~cosv-e^ucosv~sinv)+e^uz_y(e^usinv~sinv+e^ucosv~cosv)=e^{2u}z_y$.

Hence option $(A)$ is true.