Multivariable partial derivative

[RichardJ]

Homework Statement

From the transformation from polar to Cartesian coordinates, show that

\begin{equation}
\frac{\partial}{\partial x} = \cosφ \frac{\partial}{\partial r} - \frac{\sinφ}{r} \frac{\partial}{\partialφ}
\end{equation}

Homework Equations

The transformation from polar to Cartesian coordinates is assumed to be x = r\cosφ

The Attempt at a Solution

To solve the problem i tried to use the multivariable chain rule. Resulting in the following equation:

\begin{equation}
\frac{\partial}{\partial x} =\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partialφ}{\partial x}\frac{\partial}{\partial φ}
\end{equation}

Writing $r = x/\cosφ$ and $\arccos(x/r) = φ$ i tried to solve this problem. But this does not give the right answer.

Am i using the right approach? I think it is necessary to use the multivariable chain rule in some form. But the partial derivative not acting on some other function seems a bit weird to me so i am not sure how to solve this problem.

 

[Ray Vickson]

Homework Statement

From the transformation from polar to Cartesian coordinates, show that

\begin{equation}
\frac{\partial}{\partial x} = cosφ \frac{\partial}{\partial r} - \frac{sinφ}{r} \frac{\partial}{\partialφ}
\end{equation}

Homework Equations

The transformation from polar to Cartesian coordinates is assumed to be x = rcosφ

The Attempt at a Solution

To solve the problem i tried to use the multivariable chain rule. Resulting in the following equation:

\begin{equation}
\frac{\partial}{\partial x} =\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partialφ}{\partial x}\frac{\partial}{\partial φ}
\end{equation}

Writing $r = x/cosφ$ and $arccos(x/r) = φ$ i tried to solve this problem. But this does not give the right answer.

Am i using the right approach? I think it is necessary to use the multivariable chain rule in some form. But the partial derivative not acting on some other function seems a bit weird to me so i am not sure how to solve this problem.

In LaTeX, standard functions look a lot better if they are preceded by '\', so you get $\sin \phi$ instead of $sin \phi$, etc.

 

[Samy_A]

$\frac{\partial}{\partial x} =\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partialφ}{\partial x}\frac{\partial}{\partial φ}$

With that equation in mind:
$r=\sqrt{x²+y²}$
$φ=\arctan(\frac{y}{x})$ (with some subtleties).

 

[RichardJ]

Ahh, thanks a lot. That solved the problem.