(A new question of the week)
In Monday’s post about fallacies in calculus, one of them used the definition of the derivative (or rather, misused it). Today we’ll look at a short question about applying that same definition, that came in last month.
The question came from Hashem:
We all know what the arcsin(x) derivative is, but the problem is that I’m trying to obtain its derivative by applying the general definition of derivative, rather than the method of implicit differentiation, but I’m stuck and don’t know how to proceed after substituting the arcsin(x) in the definition, although I have searched the internet and found nothing. Finally, if it doesn’t work out, is that means the definition is wrong or something wrong in it? I mean the definition should apply for all functions right?
Please help me out with this problem.
It is common in calculus classes to start out by directly applying the definition of the derivative, \(\displaystyle f^\prime(x) = \lim_{h\rightarrow 0} \frac{f(x + h) – f(x)}{h}\), to several relatively simple functionslike \(f(x) = x^2\); sometimes slightly more complicated derivatives will be determined this way. Subsequently, other methods (derived from the definition) are used to find derivatives, including implicit differentiation. We rarely go back to the definition.
In this case, the usual way to obtain the derivative is to write the definition of the arcsine, \(y = \arcsin x \Leftrightarrow \sin y = x\), and differentiate both sides:
\(\displaystyle \frac{d}{dx}\sin y = \frac{d}{dx}x \Rightarrow \cos y\frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{\cos y} \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1 – x^2}}\)
Hashem wants to apply the definition directly to this function. Does that definition always work, or does it have “limits”?
Doctor Fenton replied:
Hi Hashem,
No, there is nothing wrong with the definition. The problem is how to show that the limit of the difference quotient in the definition
f(x)-f(a) lim --------- x->a x-aactually exists. With many functions, such as f(x)=Ax+B, this is straightforward, and even with more complicated functions such as f(x)=x2 or f(x)=√(x), there are algebraic identities (factoring and rationalization) that we can use to simplify the difference quotient to a form in which it is straightforward to show that the limit exists, and to evaluate it.
With the function f(x)=sin(x), these algebraic simplifications no longer work, but we still know a lot about the sine function, and can use known properties of sin(x) to show that
sin(x) lim ------- = 1 , x->0 xand with other trigonometric identities, this will show that sin(x) is differentiable.
But generally, you have to know a lot of properties of f(x) to show that the difference quotient has a limit. With functions such as arcsin(x), it is hard to determine the relationship between f(x)-f(a) and x-a, because one of the main ways to find such relationships is by using the derivative, which would make the argument circular.
Instead, there are general theorems such as the Implicit Function Theorem, the Inverse Function Theorem, and the (First) Fundamental Theorem of Calculus which can guarantee that under certain conditions a function is differentiable. If a function of interest can be shown to be differentiable using one of these theorems, you can probably construct an argument based on the argument of the theorem.
As he says, applying the definition becomes much more difficult when a function is more complicated. We know many identities that can be applied to trig functions to simplify a difference quotient and find a limit; fewer identities are familiar for the inverse trig functions, and perhaps fewer are even available at all. But in principle it should be possible to merge theorems about the derivative with the definition, so as to apply the definition in the same way it is applied in the proof of the theorem. Now he demonstrates this:
For example, you need to show that
arcsin(x)-arcsin(a) lim ---------------------- x->a x - aexists. If we let t = arcsin(x) and A = arcsin(a), then sin(t)=x and sin(A)=a, and we can write
arcsin(x) - arcsin(a) t - A ----------------------- = ----------------- , x - a sin(t) - sin(A)and the difference quotient on the right should look familiar: it is the reciprocal of the difference quotient for sin(x) at a. You need to show that if x->a, then t->A, which just says that arcsin(x) is continuous. So the limit of the left side as x->a will have the same value as the limit of the right side as t->A, and you know what that limit is. (This is basically the general argument for the Inverse Function Theorem in one dimension, restated for the specific function arcsin(x), the inverse function of sin(x).)
Can you finish the argument from here?
The Inverse Function Theorem says that, under appropriate conditions, \(\displaystyle \left(f^{-1}\right)^\prime (y) = \frac{1}{f^\prime (x)}\). That is, \(\displaystyle\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}\). It is closely related to implicit differentiation. By making a change of variables from x and a to t and A, we end up with a difference quotient whose limit can be found in the same way we do for the sine. So this derivation amounts to intermingling the proof of the derivative of the sine with a proof of the inverse function theorem.
Hashem responded:
Thank you very much Doctor Fenton, and thank you for your efforts, that was very helpful, and I completed the argument, but One final note, you said and I quote “But generally, you have to know a lot of properties of f(x) to show that the difference quotient has a limit. With functions such as arcsin(x), it is hard to determine the relationship between f(x)-f(a) and x-a“, so my question is: is it hard or impossible to determine the relationship? And is there a lack or shortage in properties of arcsin(x)? I mean, is there no alternative way in which I don’t do what you did by letting t=arcsin(x) and A=arcsin(a), because in Wikipedia I found this property of arcsin(x) subtraction, https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities, and so, Is this also don’t work out by any means?
The identities in this list are for differences of angles: \(\sin(\alpha – \beta)\), not for differences of sines, which is what \(\arcsin(\alpha – \beta)\) is, which would be needed for a fully direct derivation.
Doctor Fenton focused in his reply on the issue of proving an impossibility, and on the response to difficulty:
Kurt Goedel showed that in any axiom system which is strong enough to allow the construction of basic arithmetic, there are statements which are true but not provable in the system. When you can’t prove something, it could be because you just haven’t figured out how to prove it, or it may be impossible to prove. Until someone proves (or disproves) it, you can’t know which is true.
Computing a derivative directly from the definition can be like trying to climb a 500 meter sheer cliff, but there may be an easy path to the top if you go around to the back of the mountain. Unless you really enjoy difficult climbing, there is an easier method to get to the top, which may let you attempt more interesting climbs.
Computing some derivatives directly using the definition is a very valuable step in learning what being differentiable means, but it is not the only way, and it isn’t necessary to use the definition directly.
Thus, trying for a direct proof of something that can be proved indirectly with ease is not generally worth the struggle. But giving it a try can be educational — if you don’t keep trying too long and exhaust yourself.