What does it mean when we say $\frac{dy}{dx} = f(x)$?

Question

If we have an equation that's true for all $(x,y)$ in a set $S$, for example:
$y = 2x^2 + 3x$

We say that its derivative is:
$\frac{dy}{dx} = 4x + 3$

However, if we use the definition of the derivative on the left half $f(x) = y$, we get:
$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{y - y}{h} = 0$

This would mean that either:

• Whenever $y = 2x^2 + 3x \rightarrow 0=4x+3$, which is false
• $\frac{dy}{dx}$ doesn't always mean "the derivative of $f(x) = y$"

The same problem also happens with implicit differentiation:

$x^2 + y^2 = 1$
$2x + \frac{dy^2}{dx} = 0$
$2x + lim_{h\to0}\frac{y^2 - y^2}{h} = 0$
$2x = 0$ (not equivalent to $x^2+y^2=1$)

All I did to get that result was to use the limit definition of the derivative on $\frac{dy}{dx}$ instead of leaving it as it is and isolating it.

What am I doing wrong?

Also, by $x$ and $y$, I am referring to variables in a two-variable equation. I'm not using "y" to mean "f(x)".

Answer 1

The expression $\frac{dy}{dx}$ is read "the derivative of the variable $y$ with respect to the variable $x$". Actually, this refers to the derivative of the function that relates the variables.

If $y=2x^2+3x$ and $y=f(x)$, then $f(x)=2x^2+3x$ and $f(x+h)=2(x+h)^2+3(x+h) \neq y$. This is your mistake.

The definition is for the derivative of a function, not of a variable. $y=f(x)$ and $y=f(x+h)$ are different ways of relating the variables, and that is what causes a zero result (if both hold for all small values of $h$, this would mean that $y$ is constant with respect to $x$, so the derivative is zero).

You have the right to say that $x$ and $y$ are variables and that $y$ is not only an expression for $f(x)$, but it must be clear what function you are trying to differentiate. And remember that $y=f(x)$ is an equation relating the variables as much as $x^2+y^2=1$ is

Answer 2

Your reasoning is wrong in here:

$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{y - y}{h} = 0$

In fact, $\lim_{h\to0} f(x+h) = f(x) + h \cdot g(x)$ and it is exactly this $g(x)$ which is the value of interest and which is the derivative. In many cases, $g(x)$ can be evaluated without using calculus.

Answer 3

Using the definition,

\begin{eqnarray} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} &=& \lim_{h \to 0} \frac{2(x+h)^2 + 3(h+x) - 2x^2 + 3x}{h}\\ &=& \lim_{h\to 0} \frac{4xh + 3h + h^2}{h}\\ &=& \lim_{h\to 0} \left(4x + 3 + h \right)\\ &=& 4x + 3 \end{eqnarray}

Writing $\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{y - y}{h} = 0$ is not correct.

Since $y$ is a variable that is dependent on $x$, $y$ is function of $x$ and can be written as $y = f(x)$.