1. ## Logical equivalence

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?

2. ## Re: Logical equivalence

Originally Posted by SweatingBear
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?
1) There are functions that do not have derivatives.

2) $\displaystyle x^3+2~\&~x^3-2$ have the same derivative but are different functions.

3. ## Re: Logical equivalence

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.

4. ## Re: Logical equivalence

What if I write

$\displaystyle f(x) = x^2 + x + 1 \ \Leftrightarrow \ f'(x) = 2x + 1$

Still not valid?

5. ## Re: Logical equivalence

My lecturer advised his students to do mental workout in the morning: to take a constant and differentiate it. Next morning, he would say, take a bigger constant.

6. ## Re: Logical equivalence

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.

7. ## Re: Logical equivalence

no. if we let $\displaystyle g(x) = x^2 + x + 2$ then $\displaystyle g'(x) = 2x + 1$, so we cannot conclude from f'(x) = 2x + 1, that $\displaystyle f(x) = x^2 + 2x + 1$.

we have "=>" but not "<=" therefore not "<=>".