Now, I know that this will just become $1-\frac{2}{x^2+1}$ if I apply the derivative of arctan, but how can I calculate the derivative of this function, step by step? I am already lost at the beginning. If $f(x)=\text{arctan}(x)$ instead I would just substitute it so that $\text{tan}(y) = x$ and then use implicit differentiation, but I don't know where to start with the function in the title.
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1$\begingroup$ You can just apply the linearity of the derivative. $\endgroup$– Peter ForemanOct 10, 2019 at 22:00
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$\begingroup$ What do you mean by step? Do you mean from first principles? $\endgroup$– AllawonderOct 10, 2019 at 22:11
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$\begingroup$ Well I know that (d/dx)(tan^-1(x)) = 1/(1+x^2), but what if I didn't know that initially. How would I get from f(x) to the derivative then? This is what I mean by step, basically the manipulation that would led me there. Not sure what you meant by from first principle? $\endgroup$– NiloOct 10, 2019 at 22:16
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1$\begingroup$ If $F=f+g$ and both $f$ and $g$ are differentiable at $x$, then also $F$ is differentiable at $x$, with $F'(x)=f'(x)+g'(x)$. Since you already know the derivative of the arctangent, what's the point in computing it again? $\endgroup$– egregOct 10, 2019 at 22:49
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If you really want to apply this method then we would have $$y=x-2\arctan{(x)}$$ $$x=\tan{\left(\frac{x-y}2\right)}$$ Which we can differentiate both sides with respect to $x$ giving $$1=\sec^2{\left(\frac{x-y}2\right)}\cdot\left(\frac{1-y'}2\right)$$ $$1=(1+x^2)\cdot\left(\frac{1-y'}2\right)$$ $$\frac{1-y'}2=\frac1{1+x^2}$$ $$1-y'=\frac2{1+x^2}$$ $$y'=1-\frac2{1+x^2}$$
answered Oct 10, 2019 at 22:16
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$\begingroup$ This begs the question: how do you know that $y$ is differentiable? $\endgroup$– egregOct 10, 2019 at 22:50