# Good exercises that force you to apply the definition of the derivative, without explicitly telling you to do so?

I'd like to ask my students whether some real function is differentiable at a certain $$x_0$$. I prefer not telling them that they have to use the definition of the derivative, but to instead present a question that by its nature would require the usage of the definition. Usually I achieve this goal by giving them some piecewise defined function $$f(x)$$ which would force them to use the definition.

The problem is the following theorem which can easily be proved using the mean value theorem: if $$f(x)$$ is continuous in a neighbourhood of $$x_0$$, and differentiable in that neighbourhood except possibly at $$x_0$$ itself, then in case $$\lim_{x\to x_0} f'(x)$$ exists then $$f(x)$$ is differentiable at that point with derivative equal to that limit, and the result is also similarly correct for one sided limits and one sided derivatives.

This means that students often use this theorem either because they have seen a TA use it, or worse, because they don't see why it would require proof in the first place. In either case, this does not check their understanding of the definition (and I am left with the dilemma of how to grade these correct solutions which implicitly use a theorem which was not proven in class).

My question is: are there exercises which you must solve using the definition of the derivative and which do not easily allow a cop out such as the one above?

My thoughts are that there are 3 options-

1. A function which is only differentiable at $$x_0$$ but not in a neighbourhood of $$x_0$$. For example the function given in this question. All generalizations of this example seem to be very similar to the original one, so it is not a good exam question. There are more examples in the answer to that question, but they all rely on infinite series which have not yet been covered at this point.

2. A function for which continuity at $$x_0$$ is hard to prove. This seems unlikely, because if it's hard to prove continuity, it will also be hard to differentiate the function at that point.

3. A function which is not differentiable at all, not at $$x_0$$ nor around it. This would disallow calculating $$f'(x)$$ and then taking its limit, but then we are left with mostly boring examples and in any case the solution is always that the derivative does not exist.

I'd appreciate any ideas for such a function which forces you to differentiate "by hand", or any thoughts whatsoever on how to tackle this problem.

• Maybe the lower half (begins with "Update") of this answer? Jan 25 at 18:17
• @DaveLRenfro Thanks for this, I've never seen this before. I don't think it's directly relevant because the $\lim_{x\to x_0}f'(x)$ trick does work for it (with similar complexity to a direct calculation of the derivative). For any $x$ close enough to $0$ we have $\lfloor -x^2 \rfloor = -1$ and so $f(x)=x\lfloor -x^2 \rfloor = -x$ so $f'(x)$ is identically $-1$ around $x_0=0$ so $\lim_{x\to 0} f'(x)$ does agree with $f'(0)$. That's still a great example and I hope to remember putting it in a problem sheet.
– Snaw
Jan 29 at 16:50

Here are some example questions.

The graph of the function $$f$$ is given above. Evaluate the following limits. If the limit is infinite, write $$\infty$$ or $$-\infty$$ as appropriate. If the limit does not exist write "DNE".

\begin{align*} &\lim_{h \to 0^+} \frac{f(4+h)-f(4)}{h}\\ &\lim_{h \to 0} \frac{f(4+h)-f(4)}{h}\\ &\lim_{x \to 4^-} \frac{f(4)-f(x)}{4-x}\\ &\lim_{h \to 0^-} \frac{f(2+h)-f(2)}{h}\\ &\lim_{x \to 2^+} \frac{f(x)-f(2)}{x-2}\\ &\lim_{h \to 0} \frac{f(3+h)-f(3)}{h} \end{align*}

To successfully answer such questions, the student must have a pretty strong grasp of one sided limits, and the interpretation of difference quotients as slopes.

• +1 I really like the idea to formulate such exercises graphically. Jan 28 at 12:44
• I really like this/these exercise. They make an excellent transition from limit & continuity to the definition of the derivative. Jan 28 at 20:34

I'll argue that this will be likely not feasible for a test question.

There's several points in the calculus progression where there's a "hard bottleneck" of some sort, but once you get past that, you can leverage the new fact to easily solve previously-hard problems. One example: finding the derivative of a monomial (from the definition + binomial theorem), plus the sum rule, and then all polynomials become trivial. Second example: finding the derivative of the first trigonometric function, and then all trig functions become trivial.

Yes, ideally we want our students to understand those key proofs. No, after you've done it once, it's not really possible to put it back in the bottle and forget the new foundational fact.

So while somewhat awkward, I'll propose that to confirm understanding of those critical proofs, you just have to give an exercise that says, "prove this from the definition". Anything else will likely be far too complicated or unnatural in an exam situation (per tag in OP).

Give a function together with some properties of the function, but do not give a formula, and ask for the derivative.

An example:

Let $$f: \mathbb{R} \to \mathbb{R}$$ be a function which enjoys the following two properties:

1. $$f(t+s) = f(t)f(s)$$ for all real numbers $$t$$ and $$s$$.
2. $$\displaystyle \lim_{x \to 0} \frac{f(x) - 1}{x} = 3$$.

Prove that $$f'(x) = 3f(x)$$ for all real $$x$$.

These kinds of problems used to appear on the AP Calculus BC "essay portion" in the 60s and early 70s. I know because my AP calc teacher had us do every single AP problem going back to the inception of the AP test.

• More theoretical exercises are definitely a very good solution. Thank you for this idea.
– Snaw
Jan 29 at 16:31

This might not be what you want, but sometimes a backwards problem appears more difficult than it seems and the difficulty is immediately removed when you know and apply the definition of the derivative.

Problem: Calculate the following limit: $$\lim_{x \rightarrow a} \frac{\cos^2(3x)-\cos^2(3a)}{x- a}$$ Solution: By definition of derivative, the limit above is $$f'(a)$$ for $$f(x) = \cos^2(3x)$$ hence $$f'(a) = -6\sin(3a)\cos(3a)$$ and $$\lim_{x \rightarrow a} \frac{\cos^2(3x)-\cos^2(3a)}{x- a} = -6\sin(3a)\cos(3a).$$

• I love this idea but I'm worried that most students would solve this problem using L'Hopital's rule. This could be a great question for a midterm or any exam where L'Hopital's rule hasn't been seen yet.
– Snaw
Jan 29 at 19:52
• Using L'Hopital's rule here is pretty much the same thing as the intended solution anyway; it reduces the question to the derivative of the function at $x = a$ one way or the other. Jan 29 at 20:42
• @ChrisCunningham I think the OP’s learning goal here is to recognize and apply the definition of the derivative, not just to use derivative rules (or just to evaluate limits) for any old reason. Jan 29 at 23:27

A nice class of examples that can be used for this purpose are functions that are differentiable but not continuously differentiable (since the "$$\lim_{x \to x_0} f'(x)$$ trick" doesn't work for them). A typical example is the following exercise:

Exercise. Let $$f: \mathbb{R} \to \mathbb{R}$$ be given by $$f(x) = \begin{cases} x^2\sin(\frac{1}{x}) \quad & \text{if } x \not= 0, \\ 0 \quad & \text{if } x = 0. \end{cases}$$ Show that $$f$$ is differentiable at $$x=0$$ and compute $$f'(0)$$.

The differential quotient at $$0$$ is clearly $$0$$, so $$f'(0)$$ exists and is equal to $$0$$. However, for $$x \not= 0$$ one has $$f'(x) = 2x\sin(\frac{1}{x}) - \cos(\frac{1}{x}),$$ which does not converge as $$x \to 0$$.

• Oh, that's a very good point. I forgot about these and should've listed a 4th case for when when $\lim_{x\to x_0} f'(x)$ doesn't exist. I usually show this in class as an example of a function which is differentiable yet not continuously differentiable, and I think that modifications would result in a pretty similar exercise. Do you think it's possible to easily generate functions with this property but that are not very similar to this one? I haven't yet given this enough thought.
– Snaw
Jan 29 at 16:29
• A favorite variant is to use ${1\over2}x + x^2 \sin(1/x)$ & show that $f'(0)>0$ but $f$ is not increasing over any interval containing $0$. Jan 29 at 23:31
• @Raciquel I think these first three examples with $f(x)=(x-c)^2g(x)$ give us a function which is not twice differentiable but still continuously differentiable. If we take $g(x)=1$ at rationals, $0$ at irrationals we get a function that is not differentiable around $0$ (this is a variant of the function I linked to in #1 in the list above). Maybe there's a way to take these functions and tweak them in order that we get functions that are differentiable but not continuously so?
– Snaw
Jan 30 at 0:50
• @Raciquel The motivation is that they won't be able to use the theorem that says that if $\lim_{x\to x_0} f'(x)=L$ and $f(x)$ is continuous around $x_0$ then $f'(x_0)=L$, because then they aren't using the definition of the derivative.
– Snaw
Jan 30 at 12:52