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snusxd
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#учеба #семестр_1 #высшая_математика
$y=2^{\arctan(\sqrt{ x })}$
$y=p'(g(f(x)))\cdot g'(f(x))\cdot f'(x)$
$p'=2^{\arctan(\sqrt{ x })}\cdot \ln 2$
$g'=\dfrac{1}{1+x}$
$f'=\dfrac{1}{2\sqrt{ x }}$
$y'=2^{\arctan(\sqrt{ x })}\cdot \ln 2\cdot\dfrac{1}{1+x}\cdot\dfrac{1}{2\sqrt{ x }}=\dfrac{2^{\arctan \sqrt{ x }}\ln 2}{2\sqrt{ x }(1+x)}$