I understand why $\dfrac {\mathrm d(x^2)}{\mathrm dx} = 2x$ since we're taking the derivative of $x^2$ with respect to $x$. Or $\dfrac {\mathrm dx^2}{\mathrm dx^2} = 1$ since we're taking the derivative of $x^2$ with respect to $x^2$ as the base variable.

From the textbook:

$$\begin{align} x^2 &= u^3-1\tag1\\[5pt] \frac {\mathrm dx^2}{\mathrm du} &= 3u^2\tag2\\[5pt] dx^2&=3u^2\mathrm du\tag3\\[5pt] 2x\mathrm dx&=3u^2\mathrm du\tag4 \end{align}$$

How did the textbook go from step 3 to step 4? Specifically how does $\mathrm dx^2 = 2x\mathrm dx$?

  • 7
    $\begingroup$ Just take this as a definition of how differentials work: $$df(t) = f'(t)dt$$ $\endgroup$ – user137731 Oct 24 '16 at 3:24
  • 7
    $\begingroup$ Well, look at your first line. You have $\frac{d(x^2)}{dx}=2x$. Multiply both sides by $dx$. Isn't that what you're looking for? $\endgroup$ – Ispil Oct 24 '16 at 3:25

This is a total derivative and you appear to be using it in the form taught as implicit differentiation. The set of variables is $\{x,u\}$. (Note that when we write $x^2 = u^3 - 1$, there is no clear division between dependent and independent variables.) Then \begin{align} d(x^2) &= \frac{\partial x^2}{\partial x} \mathrm{d} x + \frac{\partial x^2}{\partial u} \mathrm{d}u \\ &= (2x) \mathrm{d}x + (0) \mathrm{d}u \\ &= 2x \mathrm{d}x \text{.} \end{align}

It probably would have been better if your book had demonstrated the use of the chain rule here: \begin{align} \frac{\mathrm{d}(x^2)}{\mathrm{d}u} &= \frac{\mathrm{d}(x^2)}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}u} \\ &= 2x \frac{\mathrm{d}x}{\mathrm{d}u} \text{,} \end{align} and then continue on by "multiplying" both sides by $\mathrm{d}u$. (Warning: there is a very good chance your "algebra brain" has just drawn an incorrect conclusion about how the products of derivatives work.)

There are a number of horrible ideas present in what you claim is in your book: $\frac{\mathrm{d}(x^2)}{\mathrm{d}u}$ is not a fraction and you cannot go around multiplying things by plain "$\mathrm{d}u$"s safely without knowing what you are doing. In particular, if taken at face value, the step of multiplying both sides of your equation by $\mathrm{d}u$ is equivalent to destroying all information in your equation by multiplying both sides by zero. What is actually meant is that a limit on the left and a limit on the right of the equals sign approach zero in some particularly helpful way that is indicated by everything else in the equation. However, this formal manipulation can be handy for solving certain types of equations. But like any formal manipulation, it can only suggest a solution; you must still verify that the suggestion is a solution.

  • $\begingroup$ Ouch; with $x^2=u^3-1$ as a given, seeing partial derivatives used like that mentally hurts! $\endgroup$ – Hurkyl Oct 24 '16 at 8:27
  • $\begingroup$ @EricTowers: While it's true that $\frac(d(x^2), du)$ $\endgroup$ – kevin cline Oct 24 '16 at 19:58

Thanks for the all hints, I figured it out.

Continuing from $(3)$:

$$\begin{align} \mathrm dx^2&=3u^2\mathrm du\tag3\\[5pt] \frac{\mathrm dx}{\mathrm dx}\mathrm dx^2&=3u^2\mathrm du\tag{3a}\\[5pt] \mathrm dx\frac{\mathrm dx^2}{\mathrm dx}&=3u^2\mathrm du\tag{3b}\\[5pt] \mathrm dx \cdot 2x&=3u^2\mathrm du\tag{3c}\\[5pt] 2x\mathrm dx&=3u^2\mathrm du\tag4 \end{align}$$

  • 2
    $\begingroup$ You can do all of those steps if you want, but it'll be a lot easier for you going forward if you just memorize the rule in my comment. In fact if you think of $d$ as an operator you can skip most of the steps in your question too. Starting from $x^2 = u^3-1$, just apply the $d$ operator and use the rule: $$d(x^2) = d(u^3-1) \implies 2xdx = 3u^2du$$ $\endgroup$ – user137731 Oct 24 '16 at 3:37
  • $\begingroup$ I know, sometimes exam questions throws me a curve ball and I have to go back to the basics to figure them out algebraically. So knowing how some formulas work helps. $\endgroup$ – A_for_ Abacus Oct 24 '16 at 3:45
  • 2
    $\begingroup$ Well, the way you're doing it isn't technically correct. It should work everywhere in calculus 1 (maybe even calculus 2, depending on your course), but derivatives aren't really fractions. So your choices are either you do something technically incorrect and that takes longer but which'll work (your method) or you do something that is correct and will also work but you won't be able to completely understand why it works yet (my method). $\endgroup$ – user137731 Oct 24 '16 at 3:47

Differentials satisfy all of the familiar derivative laws. e.g. (where $x,y$ are variables and $n$ is a constant)

$$ \mathrm{d}n = 0 $$ $$ \mathrm{d}(x+y) = \mathrm{d}x + \mathrm{d}y $$ $$ \mathrm{d}(x-y) = \mathrm{d}x - \mathrm{d}y $$ $$ \mathrm{d}(xy) = x \mathrm{d}y + y \mathrm{d}x $$ $$ \mathrm{d}\left( \frac{x}{y} \right) = \frac{y \mathrm{d}x - x \mathrm{d}y}{y^2} $$ $$ \mathrm{d}(x^n) = n x^{n-1} \mathrm{d}x$$ $$ \mathrm{d}f(x) = f'(x) \mathrm{d}x $$

Since you can apply $\mathrm{d}$ to an identity to get a new identity, the equation you were aiming for could be derived in just one step, by applying the derivative laws

$$ \mathrm{d}(x^2) = 2x \mathrm{d}x $$

$$ \mathrm{d}(u^3 - 1) = \mathrm{d}(u^3) - \mathrm{d}1 = 3 u^2 \mathrm{d}u$$

  • $\begingroup$ Interesting. So in this perspective, $\frac {dx^2}{dx^2} = \frac{2x}{2x} = 1$, essentially taking the derivative with respect to "x" in all cases where "d" of an expression occurs. $\endgroup$ – A_for_ Abacus Oct 24 '16 at 22:51
  • $\begingroup$ @A_for_Abacus: Yep -- although note division is only meaningful if the numerator is actually a multiple of the denominator (and it is in this case). And even what you did is more work than needed; $d(x^2) = d(x^2)$ is enough to deduce $\frac{d(x^2)}{d(x^2)} = 1$! $\endgroup$ – Hurkyl Oct 25 '16 at 8:54

Really, $\frac{d(x^2)}{dx} = 2x$ and $d(x^2) = 2xdx$ are just slightly different notations for the same statement. There is no multiplication going on, it just kinda looks like it.

This is a better chain of reasoning:

$x^2+c = \int{2xdx}$.

By differentiating both sides, you get

$d(x^2)+d(c) = d(\int{2xdx})$.

$d(c)$ is $0$, and $d(\int{()})$ cancels out, so

$d(x^2) = 2xdx$

And you can (hopefully) obviously extend that reasoning to other functions


Just as you said in the first line, $\frac{d(x^2)}{dx}=2x$. If you multiply both sides by $dx$, you get number 4.

  • $\begingroup$ number 4......? $\endgroup$ – A_for_ Abacus Oct 24 '16 at 3:36
  • $\begingroup$ Step 4. (It's funny that I have to write more to be able to add my comment!) $\endgroup$ – Reza Lotfalian Oct 24 '16 at 3:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.