[Math] Derivative Notation Explanation

calculusderivativesnotation

I am learning differential calculus on Khan Academy, but I am uncertain of a few things.
By the way; I understand derivatives this far: $d^{\prime}(x)$ and this: $d^{\prime}(g(x))$

I am confused mainly about Leibniz's notation.

  1. What does the "respect" mean in "derivative with respect to x" and "derivative of y with respect to x" mean?

  2. Why does $\frac d{dx}f(x)$ have only a $d$ on top? I suspect there is a hidden variable not notated.

  3. Lastly, because this question is not as important, (but can help my understanding) what does $\frac {dx}{d f(x)}$ mean? For example, what is $\frac {dx}{d \sin(x)}\left(x^2\right)$? Other examples would be helpful.

Thanks for all the help. I really don't want to wait for my senior year in high school.

Best Answer

Q1. It means exactly what it says. :-) How much does one variable change, with respect to (that is, in comparison to) another variable? For instance, if $y = 3x$, then the derivative of $y$, with respect to $x$, is $3$, because for every unit change in $x$, you get a three-unit change in $y$.

Of course, that's not at all complicated, because the function is linear. With a quadratic equation, such as $y = x^2+1$, the derivative changes, because the function is curved, and its slope changes. Its derivative is, in fact, $2x$. That means that at $x = 1$, an infinitesimally small unit change in $x$ gives a $2x = 2$ unit change in $y$. This ratio is only exact right at $x = 1$; for example, at $x = 2$, the ratio is $2x = 4$.

This expression is the limit of the ratio $\frac{\Delta y}{\Delta x}$, the change in $y$ over the change in $x$, over a small but positive interval. The limit as that interval shrinks to zero is $\frac{dy}{dx}$.

Q2. You will rarely see, at this stage, $\frac{d}{dx}$ by itself. It will be a unary prefix operator, operating on an expression such as $x^2+1$. For instance, we might write

$$ \frac{d}{dx} \left(x^2+1\right) = 2x $$

It just means the derivative of the expression that follows.

Q3. This is an unusual formulation. Ostensibly, though, it would mean the derivative of the operand with respect to $f(x)$, which you can obtain using the chain rule:

$$ \frac{dx}{df(x)} = \frac{\frac{dx}{dx}}{\frac{df(x)}{dx}} = \frac{1}{f'(x)} $$

and

$$ \frac{d}{df(x)} g(x) = \frac{\frac{dg(x)}{dx}}{\frac{df(x)}{dx}} = \frac{g'(x)}{f'(x)} $$

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