Err, no, I did this for Pete three threads and hundreds of posts ago.
If you did (I'll believe it when I see it), you failed to cite it.
So, you are still on that train, "I did not understand and I did not ask for clarification, therefore you are wrong"?
No, you are missing the point. It's not that you failed to properly define your example that I have a problem with here. It's that between when you posted the example and when you properly clarified it, you went around telling everyone else they were wrong when it was you who had posted an ambiguous example all along.
I also have a problem with you talking down to everyone, as if you were an authority on differential calculus (you're really not) when you failed to set up an example correctly, you cite examples that don't actually support your case, and even posted downright [POST=2902804]bullshit[/POST].
After you have received repeated clarifications, do you still claim that there is an error?
Again you're dodging the issue. Sure, if $$f \,=\, 3\theta \,+\, u^{2} \,+\, v$$ then $$\frac{\partial f}{\partial \theta} \,=\, 3$$. And if you'd said that in post #18 that would have been fine. But you didn't.
So, now that you have received repeated clarifications, do you understand that
$$
\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)
$$ ?
If you just say $$f \,=\, 3\theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$$, do you understand why everyone who's ever passed university calculus is going to tell you $$\frac{\partial f}{\partial \theta} \,=\, 3 \,+\, \sin(2\theta)$$?
The above would be true IF x were an independent variable. It ISN'T. There is only ONE independent variable in the example, and that is $$\theta$$.
Bullshit. You've said yourself that $$x$$ was not a function of $$\theta$$. That makes it an independent variable.