Question

Срочно нужно решение, заранее спасибо!

Answer 1

$\frac{1}{(1-6x)^{3} } =(1-6x)^{-3}=-3*(-6)*(1-6x)^{-4}$
$18* \frac{1}{(1-6x)^{4}} = \frac{18}{(1-6x)^{4}}$
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$\frac{x}{(2-3x^{2})^{2}} =(x)'*(2-3x^{2})^{-2}+((2-3x^{2})^{-2})'*x$
$1*(2-3x^{2})^{-2}+(-6x)*(2-3x^{2})^{-3}= \frac{1}{(2-3x^{2})^{2}} - \frac{6x}{(2-3x^{2})^{3}}$
$\frac{2-3x^{2}-6x}{(2-3x^{2})^{3}} =- \frac{3x^{2}+6x-2}{(2-3x^{2})^{3}}$