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If $f(x)=\frac{x}{\sqrt{1+x^2}}$, then (fof of) $(x)$ is
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The correct answer is:
$\frac{x}{\sqrt{1+3 x^2}}$
$f(x)=\frac{x}{\sqrt{1+x^2}}$
fof$=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{2 x^2+1}}$
fofof $=\frac{\frac{x}{\sqrt{2 x^2+1}}}{\sqrt{1+\frac{x^2}{2 x^2+1}}}=\frac{x}{\sqrt{1+3 x^2}}$
fof$=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{2 x^2+1}}$
fofof $=\frac{\frac{x}{\sqrt{2 x^2+1}}}{\sqrt{1+\frac{x^2}{2 x^2+1}}}=\frac{x}{\sqrt{1+3 x^2}}$
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