27

The root of the difficulty is that $x$ appears free in $f(z)$, but we are trying to "capture" it with $g(x)$, which is illegal. When we substitute $g(x)$ into $f(g(x))$, we have a variable clash: $$ f(g(\color{red} x)) = 3^{5\color{blue}x + 1} $$ The red (first) $x$ is a different variable from the blue (second) $x$. This is clearer if we rename the bound ...


26

I think it is helpful to let students know that you are looking for their thinking while problem-solving, and not just answers. Then you can ask questions like: Find all of the points on a circle of radius $8$ that have a slope of $\frac{3}{4}$. so that they can explain their problem-solving. Even if they do end up using advanced calculators, they will ...


25

There is no royal road to geometry. - Euclid Nor calculus. The essence of calculus thinking is really the limit concept. One needs to wrap one's mind around that. Formally: it's the core technique that defines derivatives and integrals. Poetically: it's the eye-of-the-needle through which you must pass to get to the next level of mathematics. There are many ...


24

I teach calculus at a community college in the U.S. (2 year college, from which many students transfer to a university). I explain limits from about day two in an informal way ("h gets infinitely close to 0"), and talk about the problems with saying infinitely close (but I keep saying it...). I tell students that our learning journey will match the ...


18

Have you thought about the fact that you’re asking this in the middle of a pandemic for which log plots are being used all over the place to visualize the growth of COVID cases? At any rate, $${d \over dt} \ln f(t) = {f’(t) \over f(t)} = \text{relative growth rate}\,.$$ Thus where the graph is roughly straight with slope $m$, we have the number of cases ...


15

We routinely get questions related to pushing more rigor in early calculus. Usually from outstanding students and based on sample of one I like it that way logic. There's a reason why things are the way they are. And that's because most students would get the opposite of a benefit pedagogically by emphasizing increased rigor in early calculus. It's not ...


14

Whenever we measure a quantity on a log scale (such as Richter, decibels, musical pitch, or a log-plot axis), we are focusing attention on relative variation in that quantity. If $y = \ln x$, we have $$\frac{dy}{dx} = \frac{1}{x}$$ and thus, for small finite changes, $$\Delta y \approx \frac{\Delta x}{x}.$$ That is, an absolute change in the logarithm ...


11

First, yes, many teachers use that term, "anti-derivative" and "integral" only when a definite integral is in use. For your examples of each, it seems to me that you offered a clear distinction between the two expressions, i.e. that the bounds is what what makes the integral definite. For me, that's where the explanation to students tends to reach a ...


11

Start with a numerical example. Say you want to find the gradient of the tangent to $y=x^2$ at $x=1$. Obviously the point itself is $(1,1)$. Pick a nearby point, say $x=1.1$. A moment with a calculator shows $y=1.21$ and the gradient of this chord is $0.21/0.1=2.1$. Now pick a closer point, $x=1.01$. We again use the calculator to find $y=1.0201$ and the ...


10

f is not a function of (only) z - f here is a function of x as well as z. I think this explanation is intelligible to a calc 1 student, and gets at the heart of the matter.


9

Note: It's possible that in the future, Wolfram Alpha will improve and be able to answer the questions in this answer, so it's best to actually try them in Wolfram Alpha first. Use questions that involve conditional statements and generic expressions. For example: If $\lim_{x\to\infty} f(x)=1$, then what is $\lim_{x\to\infty}\frac{f(x)}{x}$ equal to? ...


8

As mentioned, a probable cause is an implicit reasoning as if every operation were a homomorphism (similar to the implicit reasoning by linearity). Similar errors include $\ln(x+y) = \ln(x)+\ln(y)$, $e^{xy}=e^x e^y$, $\int f(x)g(x) dx = \int f(x) dx \int g(x) dx$, etc. Such reasoning by (very loose) analogy can be caused by not understanding that whenever ...


7

It occurs to me that maybe you can approach this from the viewpoint of approximations. Briefly, I’m thinking of the kinds of approximations they likely use in practice, and which are often given in an appendix at the back of texts, such as the following: For $x \approx 0$ we have $\frac{1}{1-x} \approx 1+x$ (multiply numerator and denominator of left side ...


7

$$ \frac{d (3^{5x+1})}{dx} = f'(g(x))g'(x)= \frac{d \left(3^{5x+1}\right)}{d(3)} \times \frac{d (3)}{dx}. $$ However $\dfrac{d (3^{5x+1})}{d(3)}$ is undefined.


7

I think maybe if you try to write down precisely what you mean by $h<<1$ you will end up writing the definition for $\lim_{h \to 0}$. From this point of view, there is really no difference between the two approaches; what you are calling the numerical approach is the same thing as the analytical approach. For example, try to compute the derivative ...


7

I admit that I'm unable to follow the proof you give as an example in your question, but am I correct when I assume that your question simply wonders how to reconcile $dy/dx$ with the fact that $dx$ approaches zero — and hence is considered zero by your students? Then I'd simply explain the limit operation visually by exploring a curve. This can be done on a ...


6

Thinking about tangent lines as linear approximations is a great way to foreshadow Taylor series. After linear approximations in calc 1, I usually give students a few prompts to get them thinking about quadratic approximations, and then I talk briefly about how it appears in calc 2.


6

I couldn't find a lot either. Suggest playing with some logarithmic properties and constructing problems based on that. E.g. pH is log10 of the hydronium ion concentration. Could ask how the pH changes with hydronium concentration addition (assume strong acid addition, to an unbuffered solution). Of course this brings in chemistry, which weirds the kids ...


6

First off - I 100% agree with Collins here that there's no shortcuts. To really understand how derivatives work, you need to learn the epsilon-delta definition. But it's my sense that you're not really shooting for that here - it sounds like what you want is a way to convince a student, not necessarily a way to prove it to them. The difference can be hard ...


5

A couple of direct applications: Showing that a power law appears on a log-log graph as a straight line with gradient equal to the exponent of the power law (although that can be done by other, probably easier, means too). Thinking about a wind turbine that's well within the turbulent regime of the atmospheric boundary layer, the average wind speed incident ...


5

Boltzmann's equation for entropy is $S=k\ln W$, and the second law of thermodynamics is all about change in entropy. Maybe this is a place to start with your quest for a practical application of the derivative of a logarithmic function.


5

The main reason we have the word "critical point" is for the first derivative test. The statement is slightly easier if you only have to say "critical point" instead of "critical point or singular point". However, students might actually remember to check for singular points if they had a different word for this case. So I ...


5

Before inventing new notation, it is very important to learn the accepted notation and teach it to your students correctly. The question contains the following claim: $$\int \cos = \sin$$ which is not a meaningful statement. It is either incorrect, or unclear. This notational issue is not a nitpick: it is a critical issue that, if it goes uncorrected, ...


5

I'll echo other responses with the same: Do NOT introduce made up notation. I've made the effort in my Calculus courses to follow your outline while avoiding any new symbols. Similar to user20311's comment, when I first cover antiderivatives, all questions are phrased as either Find an antiderivative of $\sin(x)$, Find $F(x)$ such that $\frac{d}{dx}F(x) = ...


5

This is a VERY VERY typical problem. In fact, it's a problem even for $\frac{d}{dx}3^x$, much less your example. The way I try to deal with this is one of two ways. What has to happen first? To evaluate $3^{5x+1}$, you have to evaluate $5x+1$ first. So that is the inside function in the chain rule, just like in $\sin(x^2)$ you have $x^2$ to evaluate ...


5

The term Hergert Numbers is sometimes used in my specific region of the US for the values of $x$ where $f''(x) = 0$ or $f''(x)$ is undefined. This is in reference to Rodger Hergert, an Illinois community college professor who sometime in the 1990s became very frustrated with the fact that these numbers had no good name. But it doesn't really matter what you ...


4

I'm reading the difference between your two formulae as the fact that, in the first, you have a limit and, in the second, you compute the difference quotient for a single, small, value of $h$. When we talk about the analytic derivative we usually are in the situation that $f$ has some definite, known, formula. That often allows you compute the limit ...


4

This idea is fine, and you can use the multivariable chain rule to do it this way. Say we want to differentiate $h(x) = f(x)^{g(x)}$ with respect to $x$. Notice that we can write $h$ as the composite of $p: \mathbb{R} \to \mathbb{R}^2$ defined by $p(t) = (f(t),g(t))$ with the function $E: \mathbb{R}^2 \to \mathbb{R}$ defined by $E(u,v) = u^v$. By the ...


4

There are good examples in the other answers. You can use some of them as an introductory example to the concept of derivative, then propose the case is population growth (the population could be pretty much whatever you want), where $y'(x)$ is proportional to $y(x)$. This will give them a hint of why $e^x$ is so beloved by Math teachers You can then jump ...


4

OP: "something, that is rooted in the everyday life of the students," "aged 17-18 years." Here are two ideas, using data rather than explicit functions. (1) Charts of height vs. age (in the U.S.) show that girls at age $12$ grow about $6$ cm/yr, whereas by age $17$ the derivative of growth is nearly zero. Boys are still growing about $1$ cm/yr at age $17$. ...


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