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.
What does the "respect" mean in "derivative with respect to x" and "derivative of y with respect to x" mean?
Why does $\frac d{dx}f(x)$ have only a $d$ on top? I suspect there is a hidden variable not notated.
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.
$\endgroup$ 43 Answers
$\begingroup$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
$$ \frac{d}{df(x)} g(x) = \frac{\frac{dg(x)}{dx}}{\frac{df(x)}{dx}} = \frac{g'(x)}{f'(x)} $$
$\endgroup$ $\begingroup$This is the variable for which you are differentiating. If given $\frac {\mathrm{d}y}{\mathrm{d}x}$ then you differentiate the $x$ in the expression of $y$. If given $\frac {\mathrm{d}y}{\mathrm{d}t}$ then you differentiate the $t$ in the expression of $y$.
The hidden variable is $f(x)$.
For functions of one variable the derivatives behave as if they were fractions, such that $$\frac {\mathrm{d}x}{\mathrm{d}y}=\cfrac{1}{\frac {\mathrm{d}y}{\mathrm{d}x}}$$
derivative of the function $y=y(x)$ with respect to its variable $x$. This gets useful if more than one variable is involved. E.g. have a look at this case for some value $z$ depending on variables $x$ and $y$ and varying over time, involving several variables and partial derivatives: $$ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} $$
That is not a real fraction, but the whole is read as "derivative with respect to variable x"-operator. $$ \frac{dy}{dx} = \frac{d}{dx} y = y'(x) $$ If the variable is time, Newton's "fluxion" notation with dots is still popular $$ y'(t) = \dot{y}(t) $$ Here is how the second derivative can be written $$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{d}{dx} y \right) = \left(\frac{d}{dx}\right)^2 y = (y')'(x) = y''(x) $$
I would substitute $z = f(x)$ and see how far I get, probably applying the chain rule.
