MHB Calculating $\partial f$ for $f(r, \theta) = r\cos \theta$

[OhMyMarkov]

Hello everyone, I remember this from single-variable calculus: if f(x) = 2x, then df = 2dx, right?

What if $f(r, \theta) = r\cos \theta$, how can I express $\partial f$ in terms of $r$ and $\theta$.?

Thanks!

 

[chisigma]

Hello everyone, I remember this from single-variable calculus: if f(x) = 2x, then df = 2dx, right?<br><br>What if $f(r, \theta) = r\cos \theta$, how can I express $\partial f$ in terms of $r$ and $\theta$.?<br><br>Thanks!

In Your case $f(r, \theta)$ is function of two variables, $r$ and $\theta$, so that is...

$\displaystyle df= f_{r}\ dr + f_{\theta}\ d \theta$ (1)

... where $f_{r}$ and $f_{\theta}$ are the partial derivatives of $f(*,*)$ respect to $r$ and $\theta$...

Kind regards

$\chi$ $\sigma$

 

[HallsofIvy]

Notice that you do NOT use the "curly d" here. It is df, not $\partial f$.