Suppose y is a known function. Is it then correct to write
$\displaystyle f(x) = y \ \Leftrightarrow \ f'(x) = y'$
I mean, f(x) = y is true. And if and only if f(x) = y is true, then so is f'(x) = y', right?
to put it in a different way: equivalent is not equal. we "lump stuff together" when we make equivalences.
you can define f to be equivalent to g if f'(x) = g'(x), that is a perfectly good equivalence. but it is NOT equality.
y = f(x) = 2 and y = g(x) = 0 have the same derivative, and in some sense "act the same". but clearly 2 isn't 0.
this is why we put the +C after integrals, because we don't get a unique function, we get an equivalence class of functions.
Hello, SweatingBear!
Suppose $\displaystyle y$ is a known function.
Is it then correct to write: .$\displaystyle f(x) = y \;\; \Leftrightarrow \;\; f'(x) = y'$
I mean, $\displaystyle f(x) = y$ is true.
And if and only if $\displaystyle f(x) = y$ is true, then so is $\displaystyle f'(x) = y'$, right?
I would drop the "if and only if".
The converse is not always true.
. . If $\displaystyle f'(x) = y'$, then $\displaystyle f(x)$ may be equal to $\displaystyle y.$
Example:
$\displaystyle \begin{Bmatrix}f(x) &=& x^2 - 8 \\ y &=& x^2 + 7 \end{Bmatrix}\quad\Rightarrow\quad \begin{Bmatrix} f'(x) &=& 2x \\ y' &=& 2x\end{Bmatrix}$
The derivatives are equal, but the functions are not equal.