Question

Найти производную
$y= \frac{arcsin \sqrt{x+2} }{ x^{2} } y= \sqrt{sin(2x- \frac{ \pi }{4} )}$

$y= \frac{arctg \sqrt{2x-1} }{3x}$

Answer 1

$y'= \frac{(arcSin \sqrt{x+2})'* x^{2} -arcSin \sqrt{x+2} *( x^{2} )' }{ x^{4} }=$$\frac{ \frac{1}{ \sqrt{1-(x+2)} }*( \sqrt{x+2} )'* x^{2} -2xarcSin \sqrt{x+2} }{ x^{4} }= \frac{ \frac{ x^{2} }{ \sqrt{-x-1}*2 \sqrt{x+2} }-2xarcSin \sqrt{x+2} }{ x^{4} }$$= \frac{ \frac{ x^{2} }{2 \sqrt{-( x^{2} +3x+2)} }-2xarcSin \sqrt{x+2} }{ x^{4} }$

$y'=( \sqrt{Sin(2x- \frac{ \pi }{4}) } )'= \frac{1}{2 \sqrt{Sin(2x- \frac{ \pi }{4} )} }*(Sin(2x- \frac{ \pi }{4}))'=$$= \frac{1}{2 \sqrt{Sin(2x- \frac{ \pi }{4}) } }*Cos(2x- \frac{ \pi }{4} )*(2x- \frac{ \pi }{4})'= \frac{Cos(2x- \frac{ \pi }{4}) }{ \sqrt{Sin(2x- \frac{ \pi }{4}) } }$

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