Найти производные 1 ) y=tg sqrt x / x+1 2 ) y= arctg(e^2x)+x. Created by chaikaazzz. matematika-ru

x+1 2 ) y= arctg(e^2x)+x...

August 2022 1 1 Report

Найти производные
1 ) y=tg sqrt x / x+1
2 ) y= arctg(e^2x)+x

Answers & Comments

Alexandr130398 $1)y'=(tg \frac{ \sqrt{x} }{x+1} )'= \frac{1}{cos ^{2} \frac{ \sqrt{x} }{x+1} } * (\frac{ \sqrt{x} }{x+1})'= \frac{1}{cos ^{2} \frac{ \sqrt{x} }{x+1} }* \frac{( \sqrt{x})'*(x+1)- \sqrt{x}*(x+1)' }{(x+1) ^{2} } =\\ \\ \frac{1}{cos ^{2} \frac{ \sqrt{x} }{x+1} }* \frac{ \frac{1}{2 \sqrt{x} }(x+1)- \sqrt{x} }{(x+1) ^{2} } = \frac{1}{cos ^{2} \frac{ \sqrt{x} }{x+1} }* \frac{ \frac{x+1-2x}{2 \sqrt{x} } }{(x+1) ^{2} } = \frac{1-x}{2 \sqrt{x}(x+1) ^{2}*cos ^{2} \frac{ \sqrt{x} }{x+1} }$

$2) y'= (arctg(e^{2x})+x)'= \frac{1}{1+(e^{2x}) ^{2} }*(e^{2x})' +1= \frac{1}{1+e ^{4x} } *2e ^{2x} +1= \\ \frac{2e ^{2x}}{1+e ^{4x} } +1$

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