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.

  • $\begingroup$ $\frac {d}{dx}f(x)$ means $\frac {df(x)}{dx}$.Sometimes it's just easier to write ,especially when ,instead of $f(x)$ we have a complicated expression. $\endgroup$ Commented Oct 9, 2015 at 19:05
  • 2
    $\begingroup$ Addendum to what @user254665 said: Another, rather common notation is $\frac{df}{dx}(x)$ which means the same and I like it because - in contrast to $\frac{df(x)}{dx}$ - it puts emphasis on the fact, that you should first compute the derivative (which is a function!) and then evaluate this function at point $x$. This gets more important when you need to strictly distinguish function and point of evaluation, for example a look at the chain rule: $\frac{d(f\circ g)}{dx}(x)=\frac{df}{dx}(g(x))\frac{dg}{dx}(x)$. $\endgroup$
    – Piwi
    Commented Oct 9, 2015 at 20:56
  • $\begingroup$ One Q that always comes up is why we write $d^2x/dy^2$ for the 2nd deriv.! $\endgroup$ Commented Oct 10, 2015 at 0:14
  • $\begingroup$ The phrase "with respect to $x$" makes a lot more sense in multivariable calculus where $f$ may be a function of several variables, like $f(x,y,z)$ for example. In single variable calculus, saying "with respect to $x$" sounds a bit odd and unnecessary. $\endgroup$
    – littleO
    Commented Mar 15, 2017 at 3:53

3 Answers 3


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:

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


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

  1. 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 $y$.

  2. The hidden variable is $f(x)$.

  3. For functions of one variable the derivatives behave as if they were fractions, such that $$\frac {\mathrm{d}x}{\mathrm{d}y}=\cfrac{1}{\frac {\mathrm{d}y}{\mathrm{d}x}}$$

  1. 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: $$ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} $$

  2. 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 $$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{d}{dx} y \right) = \left(\frac{d}{dx}\right)^2 y = (y')'(x) = y''(x) $$

  3. I would substitute $z = f(x)$ and see how far I get, probably applying the chain rule.

  • 1
    $\begingroup$ You're just going to scare the kid with partials. ;) $\endgroup$
    – user137731
    Commented Oct 9, 2015 at 17:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .