<s>[INST] Find $f\circ f$ if $f(t)=\dfrac{t}{(1+t^2)^{1/2}},\ \ t\in \mathbb{R}$ 
Find $f\circ f$ if $f(t)=\dfrac{t}{(1+t^2)^{1/2}},\ \ t\in \mathbb{R}$ 

$  a.)\ \dfrac{1}{(1+2t^2)^{1/2}} \\
\color{green}{ b.)\ \dfrac{t}{(1+2t^2)^{1/2}}}\\~\\
 c.)\ (1+2t^2)\\~\\
 d.)\ \text{none of these} \\$
I tried to put the value $t=1$ and after checking concluded that it's $d.)$ but the book is giving option $b.)$.
I look for a short and simple way.
I have studied maths up to 12th grade.
 [/INST] Our formula for $f$ is
$$f(\Box)=\frac{\Box}{\sqrt{1+\Box^2}}$$
With this in mind we have
\begin{align*}
f\bigl(f(t)\bigr)
&=f\left(\fbox{$f(t)$}\right) \\
&=\frac{\fbox{$f(t)$}}{\sqrt{1+\fbox{$f(t)$}^2}} \\
&=\frac{\frac{t}{\sqrt{1+t^2}}}{\sqrt{1+\left(\frac{t}{\sqrt{1+t^2}}\right)^2}} \\
&=\frac{t}{\sqrt{1+\frac{t^2}{1+t^2}}\cdot\sqrt{1+t^2}} \\
&=\frac{t}{\sqrt{\left(1+t^2\right)+\frac{t^2}{1+t^2}\cdot\left(1+t^2\right)}} \\
&=\frac{t}{\sqrt{1+t^2+t^2}} \\
&= \frac{t}{\sqrt{1+2\,t^2}}
\end{align*}
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