What's the derivative of $arctan(x - sqrt(1+x^2))$ ?

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Answer 1 · 2016-03-29T09:20:19

$(df)/(dx)=1/(2(1+x^2))$

Let us first workout derivative of $f(u)=arctanu$ .

As $g(u)=arctanu$ , $tang=u$ and differentiating we have

$sec^2g*(dg)/(dx)=1$ or $(dg)/(dx)=1/(sec^2g)=1/(1+tan^2g)=1/(1+u^2)$

Hence for $g(u)=arctanu$ , we have $(dg)/(dx)=1/(1+u^2$ ....(A)

Now to differentiate $f(x)=arctan(x-sqrt(1+x^2))$ , we will use (A) and chain rule.

Hence $(df)/(dx)=(1/(1+(x-sqrt(1+x^2))^2))*(1-1/(2sqrt(1+x^2))*2x)$ or

$(df)/(dx)=(1/(1+(x-sqrt(1+x^2))^2))*(1-x/(sqrt(1+x^2)))$

If $x=tantheta$

$(df)/(dx)=(1/(1+(tantheta-sqrt(1+tan^2theta))^2))*(1-tantheta/(sqrt(1+tan^2theta)))$ or

$(df)/(dx)=(1/(1+(tantheta-sectheta)^2))*(1-tantheta/sectheta)$ or

$(df)/(dx)=(1/(1+tan^2theta+sec^2theta-2tanthetasectheta))*((sectheta-tantheta)/sectheta)$ or

$(df)/(dx)=(1/(2sectheta(sectheta-tantheta)))*((sectheta-tantheta)/sectheta)$ or

$(df)/(dx)=(1/(2sec^2theta))=1/(2(1+tan^2theta))=1/(2(1+x^2))$

What's the derivative of arctan(x - sqrt(1+x^2))arctan(x - sqrt(1+x^2))?