Partial derivatives and functions equal to 0

Question

If I have the function (family of curves) $$F(x,y,p)=(px)^2+p=0$$ I am under the impression that $$\frac{\partial F(x,y,p)}{\partial p}=2px^2+1$$ Is not always equal to $0$.

Please could you explain this to me?

Answer 1

That's correct: it's not always equal to zero. Nor should you expect it to be.

Let's look at a simpler example, in one variable, $x$. Let $$f(x)=x$$ The equation $f(x)=x=0$ represents a set of solutions: in this case, the set $\{0\}$. Differentiating, we get $f'(x)=1=0$, which is clearly absurd, and should indicate that you can't do such a thing and expect it to make sense.

There are two possible different definitions of "equals" here. If $f$ and $g$ are identically the same function, then indeed $f'=g'$ for all inputs $x$ -- we could say $f'\equiv g'$. However, $f(x)=g(x)$ is commonly used to refer to the set of values $x$ for which both $f$ and $g$ return the same value -- for example, if $f(x)=x$ as above and $g(x)=x^2$, these functions take the same value for $x=0$ and $x=1$ while not being identical.