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

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SUMMARY

The discussion focuses on calculating the differential of the function \( f(r, \theta) = r\cos \theta \). The correct expression for the differential is given by \( df = f_{r} \, dr + f_{\theta} \, d\theta \), where \( f_{r} \) and \( f_{\theta} \) represent the partial derivatives with respect to \( r \) and \( \theta \), respectively. It is emphasized that the notation used is \( df \) instead of \( \partial f \), which is crucial for clarity in multivariable calculus.

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OhMyMarkov
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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!
 
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OhMyMarkov said:
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

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

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