Question

Найти производную y'
$y=\frac{1}{\sqrt{1-x^{2} } }arcsinx+ln\sqrt{1-x^{2} }$

Answer 1

$y = \dfrac{1}{\sqrt{1 - x^{2}}} \cdot \arcsin x + \ln\sqrt{1 - x^{2}} \\\\y' = \bigg(\dfrac{1}{\sqrt{1 - x^{2}}} \cdot \arcsin x + \ln\sqrt{1 - x^{2}} \bigg)' =\\= \bigg(\dfrac{1}{\sqrt{1 - x^{2}}} \bigg)' \cdot \arcsin x + \dfrac{1}{\sqrt{1 - x^{2}}} \cdot (\arcsin x)' + (\ln\sqrt{1 - x^{2}})' =$

$= ((1 - x^{2})^{-0,5})'\cdot \arcsin x + \dfrac{1}{\sqrt{1 - x^{2}}} \cdot \dfrac{1}{\sqrt{1 - x^{2}}} + \dfrac{1}{\sqrt{1 - x^{2}}} \cdot (\sqrt{1 - x^{2}}) ' =\\=-0,5(1 - x^{2})^{-1,5} \cdot (1 - x^{2})' \cdot \arcsin x +\dfrac{1}{1 - x^{2}} + \dfrac{1}{\sqrt{1 - x^{2}}} \cdot \dfrac{-2x}{2\sqrt{1 - x^{2}}} =\\=\dfrac{-2x}{-2\sqrt{(1 - x^{2})^{3}}}\cdot \arcsin x +\dfrac{1}{1 - x^{2}} - \dfrac{x}{1-x^{2}}} = \dfrac{x\arcsin x}{\sqrt{(1 - x^{2})^{3}}} + \dfrac{1 - x}{1 - x^{2}} =$

$=\dfrac{x\arcsin x}{\sqrt{(1 - x^{2})^{3}}} + \dfrac{1 - x}{(1 - x)(1 + x)} = \dfrac{x\arcsin x}{\sqrt{(1 - x^{2})^{3}}} + \dfrac{1}{1 + x}$

Answer 2

Ответ: во вложении Объяснение: