# Finding the derivative of function with domain empty set

I was asked to find the derivative of the function $$\sin^{-1}(e^x + 3).$$ First I was thinking of the chain rule, but something looked strange about the function. The domain of $\sin^{-1}$ is $[-1,1]$ and $e^x + 3$ is always strictly greater than $3$. I think that there is a mistake in the problem, but assuming that there isn't: what exactly is the derivative of a function like this one that "doesn't exist"? (Is it true to say that the function doesn't exist?)

• Good catch. I suppose they could be thinking about complex arguments, but my guess is that you are correct...the problem was a mistake. If you want the differentiation exercise, use $e^x-3$ instead. At least then you have a real domain of definition. – lulu Apr 29 '17 at 18:42
• this function containes complex values, so you can find no plot – Dr. Sonnhard Graubner Apr 29 '17 at 18:43
• The formal derivative of $f(x) = \sin^{-1} (e^x + 3)$ should equal its proper derivative as a function $f: \mathbf{R} \to \mathbf{C}$, right? (At least, if the right branch of the square-root function is chosen.) – Connor Harris May 2 '17 at 14:15
• In the reals, the function is said to be undefined and so is its derivative. Mistake or trap ? – Yves Daoust May 2 '17 at 14:46
• Please tell us the source of the problem. If a book, which one (and on what page)? If online, can you provide a link? The problem is most likely either a mistake on the part of the person who posed it (possibly a typo, such as writing "$\sin$" instead of "$\sinh$") or else a subtle test of students' attention to detail -- a test you've passed with flying colors! – Barry Cipra May 2 '17 at 15:03

The fundamental part of definition of a derivative is that the function must be defined at the point under consideration. The defining equation $$f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$ does not make sense unless $f(x)$ makes sense (plus you must also ensure that $f(x + h)$ makes sense for sufficiently small values of $h$). So the correct answer to the problem is that since the function in question does not exist there is no point in even considering the derivative of the function.

Some answers mention the fact that the function could be thought of a complex function of a real variable. This is nowhere specifically mentioned in the problem and if we really want to be imaginative enough, we may well consider it to be a function of a complex variable. And this changes the whole nature of the problem. Functions of a complex variable are altogether a very different beast from the usual functions of a real variable which are seen in introductory calculus courses.

Another aspect which is mentioned in other answers is about formal differentiation. This aspect is only for polynomials (and also power series as user Ethan Bolker mentions in comment) because for such functions it can be developed formally without any reference to limit definition (and the laws of derivatives remain same). The approach does not work for arbitrary functions. One may try to develop a formal approach restricting oneself to elementary functions, but then the rules of differentiation must be taken for granted as axioms and we lose theorems of differential calculus so the whole idea of formal differentiation does not sound very appealing. Unfortunately the idea that derivatives can be handled formally (without appeal to limit processes) is something which has caught on and routinely used by students in evaluation of limits via L'Hospital's Rule.

It is best to treat this problem as a typo/mistake and luckily this one is easy to spot. A slightly "difficult to spot" mistake occurred long back in an actual competitive exam in India.

• You may find formal differentiation of power series unappealing, but that operation is part of a well established and useful mathematical theory. See en.wikipedia.org/wiki/Formal_power_series – Ethan Bolker May 9 '17 at 12:15
• @EthanBolker: Sorry I missed power series. I will include it with polynomials – Paramanand Singh May 9 '17 at 12:16

The derivative exists, but as the result of a formal process. For $\sin^{-1}(e^x + 3)$ put $u=e^x+3.$ Chain rule gives us

$$\frac{d}{dx}\sin^{-1}(e^x + 3) =\frac{d}{dx}\sin^{-1}(u)= \frac{1}{\sqrt{1-u^2}}\frac{du}{dx}=\frac{e^x}{\sqrt{1-(e^x+3)^2}}.$$

The derivative map didn't even notice the domain or range of $\sin^{-1}(x),$ since in this type of computation we're simply applying an algorithm.

Another thing to notice, is that $e^x+3>3,$ as you pointed out, so the derivative isn't defined either, since the denominator will always be the square root of a negative number.

In a way, what you're touching on here is much deeper. Look at $\ln(x)$ for example. We never consider it as a composition of $x$ with $\ln(x),$ since this seems silly. However, $x$ has domain $\Bbb{R},$ and $\ln$ has domain $\Bbb{R}_{>0}.$ The functions that the derivative map $\frac{d}{dx}$ can evaluate aren't necessarily going to make sense. But as a formal process these will exist, because this is a linear map on a collection of functions which are defined in a formal manner.

• I think it will be be more accurate if you write $e^x+3 >3$ instead of $\ge$ since $e^x>0 ~\forall x \in \mathbb R$ – Jaideep Khare May 5 '17 at 6:12
• You are quite correct. Fixed. – Chickenmancer May 5 '17 at 6:57
• I have no idea what you are talking about. – zhw. May 9 '17 at 0:35
• @zhw. Could you be more specific? – Chickenmancer May 9 '17 at 1:02
• The first principle of differentiation is that the derivative of a function $f$ at point $x$ is defined as $$\dfrac{\mathop{d}}{\mathop{dx}}f(x) = \lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$$ Since the RHS contains the term $f(x)$, which in this case is not defined, then how can the derivative of $f$ (the LHS) be defined? – user443782 May 9 '17 at 12:13

Lets explore the imaginary sine function for a second.

Sine is defined to be:

$$\sin (x) = \frac{e^{ix} - e^{-ix}}{2i}$$ one can see that $$\sin (x+2\pi n) = \frac{e^{i(x+2\pi n)}-e^{-i(x+2\pi n)}}{2i} = \frac{e^{2\pi n i} e^{ix} - e^{-2\pi n i}e^{-ix}}{2i} = \frac{1\cdot e^{ix} -1\cdot e^{-ix}}{2i} = \sin x$$ So on the complex plane, there is still a periodic character for the sine function (of $2\pi$). Because of that we will limit ourself to the part of the plane with $-\pi < Re(x) \leq \pi$.

Now lets figure out when does $\sin z = a > 1$: $$\sin z = a \Rightarrow e^{iz} - e^{-iz} = 2ia \Rightarrow e^{2iz} - 2ia e^{iz} - 1 = 0$$ $$e^{iz} = \frac{2ia \pm \sqrt{-4a^2 + 4}}{2} = \frac{2ia \pm 2i\sqrt{a^2-1} }{2} = (a \pm \sqrt{a^2-1})i$$

Meaning that if $z=x+iy$ then $$e^{ix} e^{-y} = (a\pm \sqrt{a^2-1} ) e^{i\frac{\pi}{2}} \Rightarrow$$ $$x= \frac{\pi}{2} , -y = \log (a \pm \sqrt{a^2-1})$$ Both solutions are ok and equally legit.

Remember that $$(\arcsin z)' = \frac{1}{\sin'(\arcsin z)} = \frac{1}{\cos \arcsin z}$$

But, what is the cosine at each point? $$\cos z = \frac{e^{iz} + e^{-iz}}{2i}$$ First case: $$e^{iz} = (a+\sqrt{a^2-1})i \Rightarrow e^{-iz}=\frac{1}{e^{iz}} = \frac{1}{a+\sqrt{a^2-1}}\cdot -i$$

Now you can substitute it in the definition of the cosine function and get one unique solution for the derivative on the complex plane. For the second solution you can do the same.

BUT, all that is good and well when we're talking about the complex plane and $\arcsin (e^x+3)$ is defined. If we're talking about $\mathbb{R}$ , the function isn't even defined, surely you can't derive it or anything.

You can probably make sense of this as an operation on formal power series, which can be manipulated consistently and are often useful even if they don't converge.

For the power series for $\arcsin$:

Finding the power series of $\arcsin x$

Then substitute the power series for $e^x +3$, expand and differentiate term by term. The answer will be pretty ugly. It will be the formal power series for the "function" you get by naively applying the chain rule to compute the "derivative".

I doubt that this is what the person who wrote the question had in mind. I suspect either a typo, or plain thoughtlessness.

Well, it does not exist in reals, but if you replace $x \in \mathbb{R}$ for $z \in \mathbb{C}$, then you have a well defined function, and $sin(z)$ isn't bounded. It is also easy to see that the same derivative rules work for functions from (an open subset of) $\mathbb{C}$ to $\mathbb{C}$.