Differential vs Derivative

Question

I am trying to teach myself $u$-substitution in preparation for Calculus $2$, and I don't think that I quite understand the difference between the derivative and a differential. This goes on to a further ill-understanding of what $dx$ actually means in integration. For example, if I have $f(x) = x^3$, the derivative would be $\frac{df(x)}{dx} = 3x^2$.

However, the differential, if I'm not mistaken, would be $df(x) = 3x^2dx$. What is the difference between these two - the derivative and the differential?

Lastly, the integral (sorry I was having trouble formatting) $\int x^3dx$, what is $dx$ ? What is the meaning of $dx$ ?

Thanks so much!

Answer 1

I think a graph would be helpful in understanding this concepts. When you get the derivative you could think of it as the slope of the function, which in your case is x^3, at a specific (x,y). Therefore, As long as you have a point and that slope at say x=3, which is 3(3)^2=27 then you could get a tangent line to that point on the function.

On the other hand, it is actually quite nice that you try to understand what dx really means. When we integrate, in terms of the graph, we actually take the area under the curve in a way that we add up infinitely many rectangles that the sum approaches the area under the curve. You could think of the integral sign as a summation from whatever bounds you choose and f(x) as the length and dx as the width. dx is actually a very small change in x and that is what integrating precisely means. When we take infinitely many rectangles, our approximation gets better and better.

This also gives a slight hint as to why the differentiating and integrating are opposites of one another in a way that if you integrate a function and take the derivative of the integral. You actually get the function back.

Answer 2

I think that it may help to think of the definite integral as what it really is, which is a way of measuring area. In the expression $\int_a^b x^3\mathrm dx$, the integrand $x^3$ is just height, and yet our answer should be area; so, by dimensionality considerations, $\mathrm dx$ must be a length, and indeed it stands for (as Silvanus Thompson puts it in his lovely book; or perhaps it was Underwood Dudley) "a little bit of $x$".

This explanation is made most clear, as I think @ChristianBlatter was suggesting, if you think of the definition of the definite integral as a limit of Riemann sums. There $\int_a^b f(x)\mathrm dx$ is $$ \lim_{\lVert\mathscr P\rVert \to 0} \sum_{i = 0}^{n - 1} f(t_i)(x_{i + 1} - x_i), $$ where the limit is taken over all tagged partitions $\mathscr P = (a = x_0 \le t_0 \le x_1 \le t_1 \le \dotsb \le t_{n - 1} \le x_n = b)$ as $\lVert\mathscr P\rVert = \sup \{x_{i + 1} - x_i : 0 \le i < n\}$ shrinks to $0$, if the limit exists; and, if you squint, you can see $\int$ as a generalised $\sum$ (indeed Leibniz chose the operator to suggest $\int\text{um}$), and $\Delta x_i = x_{i + 1} - x_i$ as a macroscopic increment that, in the limit, becomes the infinitesimal increment $\mathrm dx$.