I have only recently began studying calculus at school, so a non-technical answer would be greatly appreciated. While I understand the techniques for differentiation and integration, I still feel as if I don't understand why they work. Part of this bewilderment stems from the notation (and the language used to describe the notation). For example,

$$ \frac{dy}{dx}(x^2+5)=2x $$

I have heard spoken aloud as "the rate of change of y of $x^2+5$ with respect to $x$ is $2x$". I am not completely clear on what "with respect to $x$" means, but I think it means that the derivative is telling you what the rate of change for each value of $x$ is. For example, when $x=5$, the gradient is $10$. If, however, you were looking at the derivative with respect to $y$, then the gradient function would tell you what the gradient is for each $y$-value.

From what I understand, $\frac{dy}{dx}$ is also just a shorthand for a more formal limit expression rather than a ratio: $$ \frac{dy}{dx}(f(x))=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} $$

However, while the notation for differentiation is somewhat intuitive, I still find the integral notation baffling:

$$ \int f(x)dx=2x $$

Why is there no "$dy$" in this notation, but there is one in the derivative notation? When the "$dx$" is adjacent to the gradient function, what does it stand for? And what does the integral sign actually mean? I feel completely stuck, so it would be helpful if someone could walk me through the notation step-by-step.

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    $\begingroup$ $$\require{cancel} \begin{align} \text{WRONG:} & \xcancel{ \frac{dy}{dx}(x^2+5)=2x} \\ {} \\ \text{RIGHT: } & \frac d {dx}(x^2+5)=2x \\ {} \\ \text{WRONG: } & \xcancel{ \frac{dy}{dx}f(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} } \\ {} \\ \text{RIGHT: } & \frac d {dx}f(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} \\ {} \\ \text{RIGHT: } & \text{If } y = f(x) \text{ then } \frac{dy}{dx} = \frac d {dx} f(x) \\ {} \end{align} $$ $\endgroup$ – Michael Hardy Jan 11 '20 at 23:26
  • $\begingroup$ You can always just use the notation $f'(x)$ for the derivative of $f$ at $x$. For example, if $f(x) = x^2 + 5$ then $f'(x) = 2x$. $\endgroup$ – littleO Jan 12 '20 at 1:06

$$ \text{If } y = f(x) \text{ then } \frac{dy}{dx} = \frac d {dx} f(x) \xcancel{{} = \frac{dy}{dx} f(x)} $$ $$ \frac{dy}{dx} = \frac{\text{infinitely small increment of $y$}}{\text{infinitely small increment of }x} $$ \begin{align} & \phantom{\frac11}\\ & \int_a^b f(x)\,dx \\[8pt] = {} & \text{sum of infinitely many} \\ & \text{infinitely small quantities} \end{align} $dx$ is an infinitely small increment of $x$.

For example, $dx$ may be an infinitely small increment of time and $f(x)$ is speed at time $x,$ so that $f(x)\,dx= \text{rate} \times\text{time}$ ${} = \text{infinitely small increment of distnace},$ so that the integral is the total distance. Or $dx$ is an infinitely small increment of depth below the surface of a swimming pool and $f(x)$ is the pressure at depth $x$ multiplied by the width of the wall at that depth, so that $f(x)\,dx$ is the infinitely small force exerted against that infinitely small portion of the wall, and then the integral is the total force.


It's perhaps useful to start into understanding what the integral notation means by understanding the derivative notation in a different way; to give an explicit example, suppose that we had a point that is constrained to lie on the parabola defined by $y=x^2$. We can imagine moving that point around and watching how its coordinates change as we move it.

The traditional notation $$\frac{dy}{dx}=2x$$ expresses that, if the point was at some position $(x,y)$ and we started moving it, its $y$ coordinate would be changing at a rate of $2x$ times the rate at which the $x$ coordinate changed. This sentence expresses a much clearer idea than the equation: we are saying two rates are related to each other. This could be better written mathematically as $$dy=2x\,dx$$ where the equality is of rates rather than of some strange $\frac{dy}{dx}$ thing. Indeed, by a rather different formalism from how $\frac{dy}{dx}$ is defined, it is possible to make this latter equation rigorously valid.

Note that you don't need to think about tracing the parabola in any particular direction - this is true whether the point is moving forwards or backwards and regardless of how fast the point is moving. The notation is also entirely egalitarian: neither $x$ nor $y$ are given special treatment. When we say "the derivative of $y$ with respect to $x$" what we mean is to transform the previous equation to tell us that $dy$ is some factor times $dx$ - or to imagine how fast $dy$ would increase if we started increasing $dx$ at a rate of $1$.

This is a bit quaint for a parabola, but to give a meatier example, consider what happens if the point were instead constrained to a circle $x^2+y^2=1$. We know that the rate of change of $x^2+y^2$ must be $0$ because this value is constant on the circle. In equations, this says that $d(x^2+y^2)=0$. However, the rate of change of that expression is clearly $2x\,dx+2y\,dy$ by the usual rules for differentiation; so one finds out the non-trivial relationship between these rates of change: $$x\,dx+y\,dy = 0$$ which says that, no matter how we move the point along this circle, the rate of change of $x$ times the current $x$ coordinate plus the rate of change of $y$ times the current $y$ coordinate is $0$. We could, of course, rearrange this to $dy=\frac{-x}y\,dx$ if we desired which now says that $dy$ is a constant multiple of $dx$ - and is often expressed as $\frac{dy}{dx}=\frac{-x}y$ and said to be the derivative of $y$ with respect to $x$.

We could also consider what happens if we just a point that is moving in the plane. We can still consider the quantities $dx$ and $dy$ to represent changes in the points position, but we would find out that they are not related to each other - we can move the point as fast as we like in one direction without affecting its movement in the other. So, despite the variables $dx$ and $dy$ making sense, we really can't write $\frac{dy}{dx}$ because there isn't a way to write $dy$ as a multiple of $dx$. However, we could have more variables - for instance, if $z=x^2+y^2$, we could note that $dz=2x\,dx+2y\,dy$ as an example of a relation. Note that this case (where there is more than one dimension of freedom) doesn't lend itself particularly well to differentiating with respect to variables, since nothing is a simple multiple of anything else, but that this doesn't stop us from considering rates of change.

While there are a lot of subtleties that show up in this notation, it suffices to think about $dx$ and $dy$ as instantaneous velocities, where an equation such as $dy=f'(x)\,dx$ states that, if the current $x$ coordinate is known, the velocity in the $y$ coordinate is some multiple of the velocity in the $x$ coordinate. This is also why the derivative tells one how to approximate the function by a linear one in a small range: if we treat $dx$ and $dy$ as changes of the function over small intervals, we lose the exact equality, but find out that the change in $y$ is close to $f'(x)$ times the change in $x$ so long as the change in $x$ is small enough; this is what the usual definition of a derivative captures.

Thinking about it this way gives some hints about how to think about the term in the integral. If we wanted to think about a simple integral such as $$\int_{0}^12x\,dx$$ we now have a term $x\,dx$ that should be thought of as a rate of change, specifying that something changes in proportion to $x$. The something here is the value of the integral. What we may think of is that we start the variable $x$ at $0$ and slowly increase it to $1$. At the start, we consider the integral to be $0=\int_0^02x\,dx$, but then update the integral based off of the relation that the rate of change of the integral ought to be $2x\,dx$ as we increase $x$.

To be explicit, maybe we try to approximate this by saying that we increase $x$ from $0$ to $1$ in one fell swoop - and maybe we'll approximate the rate of change of that integral by choosing some point $x$ in that interval and evaluating $2x\,dx$ there. Say we choose the upper endpoint $x=1$; then, the integral is suppose to change in like $2\cdot 1 \,dx$ - so it should be twice the change of $x$, giving an estimate of $2$. Of course, if we chose the lower endpoint, we'd get that the change in the integral is proportional to $0\,dx$ so should be $0$ - and if we chose points in between, we'd get answers in between.

To get a better estimate, we get closer to an instantaneous change: maybe we increase $x$ from $0$ to $1/2$ to $1$. Over the interval $[0,1/2]$ we estimate $2x\,dx$ to be around $2\cdot 1/2\,dx = dx$, so the change contributed by this interval is just change $1/2$ in $x$. Over the interval $[1/2,1]$ we esimate $2x\,dx$ to be around $2\cdot 1\,dx = 2\,dx$, so this interval contributes twice the change of $x$ - which is $1$. So we estimate the integral as $1\cdot 1/2 + 2\cdot 1/2 = 3/2$. Note how each term is a value ($2x$) times a change ($dx$), which is where the notation $2x\,dx$ comes from: we want to take all the little changes, multiply them by some constant, and then sum them up to estimate how big the integral should have become - and we get this answer by dividing into smaller and smaller intervals and repeating this process.

A more complicated example that really illustrates this is the fact that integration is not restricted to intervals on the real line: we could consider a scenario such as "We have a point traveling along the parabola $y=x^2$ between the points $(0,0)$ and $(1,1)$. There is a wind blowing on the point with a force of $(-F_1(x,y),-F_2(x,y))$ at each point along the parabola. How much energy is required to push the point along this path?" The only physical thing we need to know is that the energy to push a point from a point, say $(x_0,y_0)$ to a point $(x_1,y_1)$ against a force of $(-a,-b)$ is $a(x_1-x_0)+b(y_1-y_0)$ - that is, it is a product of the force with the change in position. Then, we can say that, at each moment, the change in the total amount of energy is $F_1(x,y)\,dx+F_2(x,y)\,dy$, since we can imagine that along any short enough distance, the force remains relatively constant, and the energy used on that segment is related to the change of the two coordinates. Thus, if we somehow combine the information about the path and the endpoints into a symbol $\gamma$, the answer would be $$\int_{\gamma}F_1(x,y)\,dx + F_2(x,y)\,dy$$ which really means "break up the path into small pieces and sum a quantity over each piece, taking into account both how $x$ changed and how $y$ changed". The point of this example is that these $dx$ and $dy$ terms are quite flexible: they really do refer to a change in a variable and can directly translate physical meaning into mathematical meaning.

  • $\begingroup$ Using differentials does make the derivative notation more intuitive. In your post, you said that it is possible to make the equation $dy=2xdx$ rigorously valid. How would you do this? $\endgroup$ – Joe Mar 16 at 16:53
  • $\begingroup$ My trouble understanding how they work is perhaps illustrated by the equation $d(x^2+y^2)=0$. It makes sense intuitively that since $x^2+y^2$ is a constant, that $d(x^2+y^2)$ must be $0$. However, I'm still unsure about what $d(x^2+y^2)$ actually is. $\endgroup$ – Joe Mar 18 at 21:08
  • $\begingroup$ @Joe The short answer is "it's a differential form". Loosely speaking, to define these one first imagines picking some point $p$ on the circle $S^1$ defined by $x^2+y^2=1$ and considering differentiable curves $\gamma:\mathbb R\rightarrow S^1$ so that $\gamma(0)=p$. Differential forms are functions that accept such a $\gamma$ as input (and work at every point $p$) - $dx$, for instance, consumes $\gamma$ and returns the $x$ component of $\gamma'(0)$. $d(x^2+y^2)$ is another form - it returns $(f\circ \gamma)'(0)$ where $f(x,y)=x^2+y^2$. (cont.) $\endgroup$ – Milo Brandt Mar 18 at 22:50
  • $\begingroup$ $y\,dy$ takes $\gamma$ and returns $y$ times the $y$ component of $\gamma'(0)$. The trick with these things is that it is always true, for instance, that, if we let $\pi_1(x,y)=x$ and $\pi_2(y)=y$ we have $$(f\circ\gamma)'(t) = 2(\pi_1\circ \gamma)(t)\cdot (\pi_1\circ\gamma)'(t) + 2(\pi_2\circ\gamma)(t)\cdot (\pi_2\circ\gamma)'(t)$$ for any $\gamma$ and with $f(x,y)=x^2+y^2$ - that's the most elementary meaning of these things. But rigorously defining them is, unfortunately, a rather difficult thing to do (roughly speaking - differential geometry is the field that studies these things). $\endgroup$ – Milo Brandt Mar 18 at 22:52

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