If we rewrite $\displaystyle \frac {d} {dx} \cot(x)$ as $\displaystyle \frac {d} {dx} \frac {1} {\tan(x)}$ and then apply the quotient rule, we get to $\displaystyle \frac {\tan(x)\frac{d}{dx}1-1\frac{d}{dx}\tan(x)} {\tan^2(x)}$ and then $\displaystyle \frac {-\sec^2(x)} {\tan^2(x)} = -\frac {1} {\cos^2(x)} \cdot \frac{\cos^2(x)} {\sin^2(x)} = -\csc^2(x)$
My question is that will this proof be valid for $\displaystyle \frac {\pi} {2}$? The derivative of $\tan(x)$ is $\sec^2(x)$ only for angles for which $\tan(x)$ is defined. $\tan(x)$ is undefined for $\frac {\pi} {2}$, so in the above quotient rule, when it is claimed that $\frac {d} {dx} \tan(x) = \
sec^2(x)$, that comes with the caveat that $x \neq \frac {\pi} {2}$ (as well as $\frac {3\pi} {2}$ and their co-terminals).
Now $\frac {\pi} {2}$ is in the domain of $\cot(x)$, but I don't think that the proof in the opening paragraph holds for $\frac {\pi} {2}$
If I go by the first principle i.e.
$$\displaystyle \lim_{h \to 0} \frac {cot(x+h)-cot(h)} {h}$$, then I get a proof which works for all angles in the domain of $\cot(x)$ including $\frac {\pi} {2}$
Similarly, if I apply the quotient rule on $\displaystyle \frac {d} {dx} \cot(x) = \frac {d} {dx} \frac {\cos(x)} {\sin(x)}$, I don't run into any problems with $\frac {\pi} {2}$
So, I am just curious if the proof outlined in the first para is applicable for $\frac {\pi} {2}$. I am not so convinced, but wherever I have seen that approach on the internet, I haven't seen anyone make a note or caveat that this approach may have some issues.
Thanks.
Best Answer
If you use a definition of $\cot$ that is not defined for $\frac \pi 2$ like $\cot x=\frac 1 {\tan x}$, then the proof that you make will not prove the derivative for $x=\frac \pi 2$. Meanwhile, if you do not make that assumption and use the general properties of $\cot$ like the $\cot$ addition identity or $\cot x=\frac{\cos x}{\sin x}$, then you will have a more robust proof.
People probably use $\cot x=\frac 1 {\tan x}$ because it's easy and leads to the right answer, even if it does not prove the derivative for $x=\frac \pi 2$.