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Differentiate $$ \text{arccot} \frac{1-x}{1+x} $$ with respect to $x$

After putting $x= \cos \theta$, I got $$\text{arccot} \left(\tan^2 \frac{\theta}{2}\right)$$ Then how do I reach the answer?
Thank you in advance

Brian M. Scott
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chndn
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    Why did you make a substitution that is not just unnecessary but actually counterproductive?! You want the derivative with respect to $x$, so you do not want to make $x$ disappear. Just differentiate the thing as it stands. Or if you want to do something genuinely useful (because it makes the differentiation a little easier), start by rewriting $\frac{1-x}{1+x}$ as $-1+\frac2{1+x}$. – Brian M. Scott Aug 25 '13 at 07:39
  • @BrianM.Scott, why is it that you preach against differentiation by substitution? I am curious to know what is the reason behind it, since if a man of your level of experience says against that, then there must be something in it! I'd like to what it is. (Btw, I am not at all trying to sound impolite if the words suggest so!) In fact, I myself used to do brute chain rule when I first learnt it. However when I came to know about the apparent ease we gain via substitution, I switched over to using it whenever possible. – Parth Thakkar Aug 25 '13 at 09:15
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    @Parth: If you know how to differentiate mechanically and are comfortable doing so, then there’s no harm in learning more sophisticated techniques. For someone who is having trouble with the basics, however, I think that adding the extra conceptual layer of substitution is just inviting further confusion: it’s a separate concept, and thus just one more thing to misunderstand. – Brian M. Scott Aug 25 '13 at 18:12
  • Okay! Got it. While answering in future, I'll keep this in mind. – Parth Thakkar Aug 25 '13 at 18:13

3 Answers3

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Method $1:$

HINT:

Putting $x=\tan \theta \implies \theta=\arctan x$

$$\frac{1-x}{1+x}=\frac{1-\tan\theta}{1+\tan\theta}=\frac{\tan\frac\pi4-\tan\theta}{1+\tan\theta\tan \frac\pi4}\text{ as } \tan\frac\pi4=1$$ $$=\tan\left(\frac\pi4-\theta\right)=\cot\{\frac\pi2-\left(\frac\pi4-\theta\right)\}=\cot\left(\frac\pi4+\theta\right)=\cot\left(\frac\pi4+\arctan x\right)$$

Then, $$ \text{arccot} \left(\frac{1-x}{1+x}\right) =?$$

Method $2:$

Let $$ y= \text{arccot} \frac{1-x}{1+x}\implies \cot y=\frac{1-x}{1+x} $$

Applying Componendo and dividendo, $$x=\frac{1-\cot y}{1+\cot y}=\frac{\tan y-1}{1+\tan y} (\text{ multiplying the numerator & the denominator by}\tan y)$$

$$\implies x=\frac{\tan y-\tan\frac\pi4}{1+\tan\frac\pi4\tan y}=\tan\left(y-\frac\pi4\right)\text{ using }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$

Differentiating wrt $y,$ $$ \frac{dx}{dy}=\frac{d \tan\left(y-\frac\pi4\right)}{d(y-\frac\pi4)}\cdot\frac {d(y-\frac\pi4)}{dy}=\sec^2\left(y-\frac\pi4\right)=1+\tan^2\left(y-\frac\pi4\right)=1+x^2$$

Now use this

Community
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lab bhattacharjee
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I think this helps: www.youtube.com/watch?v=xO8PZdQOpyg to calculate $\frac{df}{dv}$ if applying the chain rule with:

$$v=\frac{x+1}{x-1}$$ and $$f=\text{arccot}(v)$$ and $$\frac{df}{dx}=\frac{df}{dv}\frac{dv}{dx}$$

The calculation of $\frac{dv}{dx}$ is left to your exercise.

al-Hwarizmi
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Seirios Aug 25 '13 at 08:11
  • @Seirios the answer includes no clarification or critique to the author. It contains the response. – al-Hwarizmi Aug 25 '13 at 08:26
  • @T.Bongers I do agree with the fact that link only answers are a problem when link changes. In this case, as it was a video, I was quite in doubt how to move things. So the answer includes the key matter that chain rule needs to be applied. I ubdated to be clearer the differential of arccot is rather minor point via video. – al-Hwarizmi Aug 25 '13 at 08:29
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Let $u = \frac{1 - x}{1 + x}$. We know that $$\frac{d}{du} arc\cot{u} = \frac{-1}{1 + u^{2}}.$$

Using chain rule, we will get $$\frac{d}{dx} arc\cot\frac{1-x}{1+x} = \frac{d}{du} arc\cot{u}\cdot\frac{d}{dx}u$$

$$= \frac{-1}{1 + \frac{(1 - x)^{2}}{(1 + x)^2}}\frac{-2}{(1 + x)^{2}}$$

$$= \frac{1}{1 + x^{2}.}$$

Petch Puttichai
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