Question

Ignore me I am just practicing something I haven't done for a while (including a latex practice too).

Answer 1

Derivative of \(\large\color{black}{ \displaystyle\left|\tan^{-1}(\sqrt{x})\right| }\)

at first,
\(\large\color{black}{ \displaystyle y=\tan^{-1}(\sqrt{x}) }\)
\(\large\color{black}{ \displaystyle \sqrt{x} =\tan (y) }\)
derivative on both sides
\(\large\color{black}{ \displaystyle \frac{1}{2\sqrt{x} }=\frac{dy}{dx}~\sec^2(y) }\)

\(\large\color{black}{ \displaystyle \frac{1}{2\sqrt{x} }=\frac{dy}{dx}~\left( \tan^2y+1\right) }\)

\(\large\color{black}{ \displaystyle \frac{1}{2\sqrt{x} }=\frac{dy}{dx}~\left( x+1\right) }\)

\(\large\color{black}{ \displaystyle \frac{1 }{2\left( x+1\right)\sqrt{x} }=\frac{dy}{dx} }\)

So I got the chain,
\(\large\color{black}{ \displaystyle \frac{dy}{dx} \left(\tan^{-1}\sqrt{x}\right)=\frac{1 }{2\left( x+1\right)\sqrt{x} }}\)


derivation of derivative of an absolute value
definition of absolute value. \(\large\color{black}{ \displaystyle\left|f(x)\right| =\sqrt{~[~f(x)~]^2~} }\)

derivative of any absolute value function
\(\large\color{black}{ \displaystyle\frac{d}{dx}\left(~\sqrt{~[~f(x)~]^2~}~\right)= \frac{1}{2\sqrt{~[~f(x)~]^2~}}\times 2[f(x)]^{2-1}\times f'(x) }\)

\(\large\color{blue}{ \displaystyle\frac{d}{dx}\left(~\sqrt{~[~f(x)~]^2~}~\right)= \frac{f(x)}{|f(x)|}f'(x)}\)

Answer 2

\(\large\color{black}{ \displaystyle\frac{d}{dx}\left(~\left|\tan^{-1}(\sqrt{x})\right|~\right)~\\[1.5em]~=\displaystyle \frac{\tan^{-1}\sqrt{x} }{2~|\tan^{-1}\sqrt{x}|~(x+1)~\sqrt{x}} }\)