I'm new to calculus and have been taught that $\displaystyle \frac{dy}{dx}$ is the rate of change of y with respect to x. Does $\displaystyle \frac{dy}{dx}$ show how much the variable y changes as x changes? Is there more to understanding this part of calculus, I feel as I'm missing the fundamentals behind differential calculus.

Also one thing that I didn't understand is when doing u-substitution integration if we let $u=2x +1$ for example, sometimes I see $\displaystyle du = 2dx \therefore dx = \frac{du}{2}$. What is this known as and why does it work? My teacher school finds $\displaystyle \frac{du}{dx}$ and rearranges this to make $dx$ the subject. Is this an incorrect practice? I have searched for this on here and cannot find a definite answer.

What is the correct notation to be used? When differentiating y = f(x) are we always operating on y as in $\displaystyle \frac{d}{dx} (y)$ = $\displaystyle \frac{dy}{dx}$. If you differentiate x^2 w.r.t x as in $\displaystyle \frac{d}{dx}x^2$, are you finding how much x^2 changes as x changes e.g if $x = 1, x^2 = 1, x = 2, x^2 = 4, x = 3, x^2 = 9$ so $x^2$ is 2 times the value of x? If we have something like $y^3$, what does it mean to differentiate $y^3$ with respect to x as in $\displaystyle \frac{d}{dx} y^3$ and how is it done?

thanks, I have been looking for the solutions to my problems for quite a while but cannot find an answer that leaves me satisfied. Sorry if questions likethese are not to be asked here.

  • 1
    $\begingroup$ +1 these are mature questions for a calculus student to be asking $\endgroup$
    – Jonathan
    Commented Aug 25, 2013 at 15:59

4 Answers 4


You are asking some excellent questions! First, you are correct in assessing the notation $\frac{dy}{dx}$ as the rate of change of $y$ with respect to $x$. It is important to keep in mind, however, that the derivative gives you an instantaneous rate of change. In basic algebra, we talk about lines and their slopes; the slope allows us to measure the rate of change of $y$ w.r.t $x$. If we now talk about more general functions -- not necessarily lines -- such as $y=f(x)=x^2$, we can formally define the derivative to be the limit $$\frac{df}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$ But this definition is essentially just the idea of slope! You take a $y$-value $y=f(x)$ and another $y$-value $y'=f(x+h)$ a little ways away, take the difference (this is your $\Delta y$) and do the same for the difference in $x$, $\Delta x=h$, and then divide. This gives you the slope of the secant line that passes through the curve $y=f(x)$ at $x$ and $x+h$. Now, the limit out front is just asking us to visualize the two points $y$ and $y'$ to be very close to each other -- the secant line thus becomes a tangent line, and instead of writing $\frac{\Delta y}{\Delta x}$, we now write $\frac{dy}{dx}$, for the slope of the line tangent to the curve at any $x$. Hence the derivative $\frac{dy}{dx}$ is a function, which when evaluated at $x$, yields the slope of the line tangent to the curve $y=f(x)$ at the point $x$.

In this sense, the derivative actually gives you information about the instantaneous rate of change of a function. Imagine the function $y=f(x)$ as the trajectory of a baseball. If we removed the limit out front from above (i.e. just computing some $\frac{\Delta y}{\Delta x}$) we'd have computed the slope of a secant line, i.e. the average velocity of the baseball during the time interval $(x,x+h)$. In taking the limit, however, we make this interval smaller and smaller until we're talking about the actual speed of the baseball at some time $x$. I hope this gives you some intuition how a derivative extends the concept of slope/rate-of-change.

As AlexR mentioned above, the $u$-substitutions we make when computing integrals are actually related to the chain rule, for which is there is a perfectly rigorous proof. You are justified in feeling uneasy about writing things like $du=2dx$, but rest assured that this is simply for notational convenience. You can think of this more rigorously as AlexR describes (or, in fact, if you learn differential geometry, the notation can be made rigorous in terms of differential forms).

  • $\begingroup$ Is it appropriate to understand that when finding the derivative, as $dx \to 0$, $dy \to 0$. Since the gradient of a tangent at a point is equal to the rise/run (difference in y/difference in x) then this becomes $$\lim_{h\to 0}\frac{dy}{dx}$$ and since both dx and dy are approaching 0 they're said to become infinitesimally small so the gradient at that point is therefore a ratio of two infinitesimals? So the procedure my teacher used is correct assuming this understanding? $\endgroup$
    – salman
    Commented Aug 25, 2013 at 16:52
  • $\begingroup$ @user90771: If $dx$ is an infinitesimal $x$-increment and $dy$ the corresponding $y$-increment (of the dependent variable $y$), then the ratio $\frac{dy}{dx}$ will be infinitely close to the derivative (or gradient, as you call it). What you are describing is the correct intuition, but to formalize the procedure one usually uses the standard part. The limit can then be defined as the standard part of the ratio of infinitesimals. $\endgroup$ Commented Aug 25, 2013 at 17:01
  • $\begingroup$ Yes, that is the correct intuitive picture of what is happening. To summarize what you're saying in slightly tighter notation: you can compute rise/run ($\Delta y/\Delta x$), but if you take a limit $$\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$$ you no longer are talking about rise/run. You're now (loosely speaking) thinking of infinitesimal changes, which is why we use the notation $dy/dx$ and not $\Delta y/\Delta x$. You can treat these as infinitesimals on their own (just $dy$ or just $dx$), and you will get the correct answer, but really what is happening is just the chain rule. $\endgroup$ Commented Aug 25, 2013 at 17:03
  • $\begingroup$ @user72694 I havent studied hyperreals or much on that wikipedia page (and I will not be studying maths to university level). Do you think it's more appropriate for me to stop worrying about intuition and focus more of problem-solving or learning more topics in calculus? Sometimes I wish to learn of things from first principles (such as the origins of e) but if this requires knowledge which is out of scope it may not be a feasible option and maybe a waste of time. $\endgroup$
    – salman
    Commented Aug 25, 2013 at 17:17
  • $\begingroup$ I think it's always worth spending time familiarizing yourself with the intuition behind mathematical concepts, even (especially) if you don't have the time to delve into the rigorous details. Learning things from first principles is always enlightening, but really what is important is understanding why certain mathematical statements are plausible -- why things should be the way they should be. Having a loose sketch of the math in your head is useful when it comes to problem solving. If there's a technical detail you don't quite remember from calculus, you can always just look it up again. $\endgroup$ Commented Aug 25, 2013 at 17:23

While the other answers are precise and to the point, the OP may be looking for motivation for the calculational practices he observed in the classroom. To address specifically his question:

"one thing that I didn't understand is when doing u-substitution integration if we let $u=2x +1$ for example, sometimes I see $\displaystyle du = 2dx \therefore dx = \frac{du}{2}$. What is this known as and why does it work? My teacher school finds $\displaystyle \frac{du}{dx}$ and rearranges this to make $dx$ the subject. Is this an incorrect practice? I have searched for this on here and cannot find a definite answer."

I would point out that the proper justification of your teacher's procedure involves infinitesimals. In this approach, $dx$ and $du$ are infinitesimals and the expression $\frac{du}{dx}$ is a ratio. While there are some details to be ironed out, you should keep in mind that the mainstream approach adopted at a majority of universities avoids infinitesimals and formulates the calculus in terms of the real numbers and epsilon, delta procedures. In this approach it is difficult to account for $\frac{du}{dx}$ as a true ratio in a meaningful way. While you will have to learn the calculus at the university the way it is taught there, it is usually helpful to think of the procedures of the calculus in terms of infinitesimals.

A revealing study of this phenomenon from the educational viewpoint is the following article by R. Ely:

Ely, Robert: Nonstandard student conceptions about infinitesimal and infinite numbers. Journal for Research in Mathematics Education 41 (2010), no. 2, 117-146. See http://www.nctm.org/publications/article.aspx?id=26196 and http://u.cs.biu.ac.il/~katzmik/ely10.pdf


$\frac{dy}{dx}$ is the rate of change of with respect to $x$. I will try to explain what this mean.

For example let us take $y=x^2$ And we are going to find out $\frac{d}{dx}(x^2)$.In other words we are going to find out rate of change of $x^2$ with respect to $x$. In other words we are going to find out how $x^2$ is changing when we introduce a change in $x$.

So, we introduce a change in $x$, say by an amount $h$, so as to make it $x+h$.Now instead of $x^2$ we have $(x+h)^2$. So how much change has been added to $x^2$?.

It is $$(x+h)^2-x^2=2xh+h^2$$ This is the amount change in $x^2$ when we introduced a change of $h$ in $x$.So what is the rate? It is $$\frac{\text{change in }x^2}{\text{change in }x}=\frac{2xh+h^2}{h}=2x+h$$ Now we note that the rate is dependent on how much change we are introducing.Actually we are not interested in large amount of change. What we want to find is an instantaneous rate of change, we want to to make the change or $h$ as small as possible,as close to zero as possible, but obviously not zero(then there would be no change!). So that $2x+h$ is is getting closer to $2x$ as $h$ gets closer to zero.

We formally write this as $$\underset {h\rightarrow 0}{lim}\text{ } 2x+h=2x$$ .

Now the formal definition of derivative, $$\frac{d}{dx}(f(x))=\underset {h\rightarrow 0}{lim}\frac{f(x+h)-f(x)}{h} $$or $$\frac{dy}{dx}=\underset{\Delta x\rightarrow 0}{lim}\frac{\Delta y}{\Delta x}$$ where $\Delta y$ is the change in $y$ as we change $x$ by $\Delta x$ .

So as $\Delta x$ gets closer to zero $\Delta y$ is also getting zero.Now we may view the derivative as ratio of two infinitesimals. This is the reason why, for instance when $\frac{dy}{dx}=2$ ,we write $dy=2dx$ . This is often useful for practical point of view. But it is not very mathematically rigorous, but one need not be worried about it most of the cases.

It must be clear now that derivative involves concept of limits, and I have explained it in a loose way.But that is okay to understand and to have a feel of what derivative is.

Now coming to your question of differentiating $y^3$ with respect to $x$, Since $y$ is a function of $x$ $y^3$ is also a function of $x$. So it should be possible to do the differentiation using the definition of derivative above.Now if you know $\frac{dy}{dx}$ there are rules known as chain rule to compute the derivative of some function of y with respect to x.But everything is basically derived from the base definition. So just for the sake of completeness $$\frac{dy^3}{dx}=\frac{d(y^3)}{dy}\frac{dy}{dx}=3y^2\frac{dy}{dx}$$

$\frac{dy}{dx}$ is the same as $\frac{d}{dx}(y)$. Also when you write $du=2xdx$ what it means is that the infinitesimal change in $u$, when you make an infinitesimal change in $x$ is 2 times $x$ times the infinitesimal change in $x$ .

I may add some more details.Say $y$ is some arbitrary function of function of $x$ ($y=f(x)$).When make change $x$ to $x+\Delta x$, f(x) will change in this way $$f(x+\Delta x)=f(x)+(\text{some expression in x})\Delta x +(\text{something}){\Delta x}^2+(\text{something else }){\Delta x}^3+......$$

You will see that by using the definition of derivative the coefficient of $\Delta x$ is the derivative of $f(x)$ with respect to $x$. Now $dy$ is ($(f(x+\Delta x)-f(x)$) as $\Delta x$ tends to zero . This can be seen to be equal to $(\text{derivative})\times \Delta x+ \text{other terms}$ as $\Delta x$ tends to zero. What we are doing is almost like ignoring the terms involving higher powers of $\Delta x$ .

  • $\begingroup$ @user90771 I have added some more details to the answer to address your further concerns. hope it helps $\endgroup$
    – jaseem
    Commented Aug 25, 2013 at 17:52
  • $\begingroup$ hey can I also ask, for $$\frac{d}{dx}y^3 = \frac{d}{dy}y^3 \frac{d}{dx}y$$ is it correct to think that the dy in the denominator and numerator cancel? $\endgroup$
    – salman
    Commented Aug 25, 2013 at 19:07
  • $\begingroup$ @user90771 It is not correct think that way. But it's easy to think that way to solve most of the problems. $\endgroup$
    – jaseem
    Commented Aug 26, 2013 at 9:18
  • $\begingroup$ @user90771 Letting $u=y^3$ what you are asking is whether it is correct to think of the identity $\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}$ (chain rule) as "cancellation" of $dy$ in numerator and denominator on the right hand side. Again thinking in terms of infinitesimals, I would say that the intuition is correct though some technical details need to be worked out (see my remarks above). For second derivatives, the distinction between dependent variable and independent variable becomes more important and one can't cancel as easily. $\endgroup$ Commented Aug 26, 2013 at 13:03
  • $\begingroup$ @user72694 yeah that is what I meant. Is this a "bad" way to look at this issue? Is there a definite answer about the infinitesimals cancelling? And what exactly do you mean by the second derivative? $\endgroup$
    – salman
    Commented Aug 26, 2013 at 17:50

$$\frac{d}{dx} y(x)^3 = 3 (y(x))^2 \cdot y'(x)$$ by the chain rule of differentiation. The notation $du = 2dx$ is not mathematically clean. Actually you use in Substitution: $$\int_a^b y(x) dx = \int_{\phi^{-1}(a)}^{\phi^{-1}(b)} y(\phi(u)) \cdot \phi'(u) du$$ And you give $u = 2x + 1 \leadsto \phi(u) = \frac{u-1}{2}$ and $\phi'(u) = \frac{1}{2}$

  • $\begingroup$ I, for one, find the notation $du=2dx$ mathematically perfectly clean (differentials of functions do belong to elementary calculus) $\endgroup$
    – user8268
    Commented Aug 25, 2013 at 16:14
  • $\begingroup$ @user8268 $u$ and $x$ are variables whereas $du, dx$ are associated Lebesgue-measures. Thus $du = 2 dx$ means $$\forall A \in \mathcal{B}: \lambda_u(A) = 2 \lambda_x(A)$$ which is not exactly "clean"... $\endgroup$
    – AlexR
    Commented Aug 25, 2013 at 16:19
  • $\begingroup$ notation is overloaded (and thus perhaps not clean). I meant differentials of (differentiable) functions, which is far more elementary than measures. $\endgroup$
    – user8268
    Commented Aug 25, 2013 at 16:23
  • $\begingroup$ Then again $du$ and $dx$ may not be separated as both are no functions, but $\displaystyle \frac{du}{dx}$ is a differentiation of a function $u = u(x)$. $\endgroup$
    – AlexR
    Commented Aug 25, 2013 at 16:27
  • 1
    $\begingroup$ The linear function on WHAT domain? See the Problem here? A linear function on the tangent space is not exactly what students are taught in calculus ;-) $\endgroup$
    – AlexR
    Commented Aug 25, 2013 at 16:33

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