Question

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$1)y = \sqrt{9 - 2x {}^{2} } \: \: 2)y = \frac{1}{ \sqrt{3x + 1} }$
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Answer 1

$1) y=f(x), y'=f'(a)*a'\\a'=(9-2x^2)'=-4x\\y'=\frac{-4x}{2\sqrt{9-2x^2} } =\frac{-2x}{\sqrt{9-2x^2} } \\\\2)y=\frac{1}{\sqrt{3x+1}}= (\sqrt{3x+1} )^{-1}=(3x+1)^{-\frac{1}{2} }\\a=3x+1, a'=3\\y'=-\frac{3}{2\sqrt{(3x+1)^3} }$