x^2)-5x...

June 2022 1 8 Report

решите пожалйста производные 1)f(x)=(sin^2)x-(cos^2)x
2)y=(sin2x-cos2x)/(sin2x+cos2x)
3)sqrt(x)^5 arcsinx-(tg/x^2)
4)(1/sqrt(x^3)^4)+(3/x^2)-5x

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NNNLLL54

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$y= \frac{sin2x-cos2x }{sin2x+cos2x}\\\\y'= \frac{2cos2x+2sin2x}{2cos2x-2sin2x}= \frac{cos2x+sin2x}{cos2x-sin2x}\\\\\\y= \sqrt{x^5} \cdot arcsin(tgx^2)\\\\y'=\frac{5}{2}x^{\frac{3}{2}}\cdot arcsin(tgx^2)+\sqrt{x^5}\cdot \frac{1}{\sqrt{1-tg^2(x^2)}}\cdot \frac{1}{cos^2(x^2)}\cdot 2x\\\\\\y=(\frac{1}{\sqrt{x^3}})^4+\frac{3}{x^2}-5x\\\\y'=4(\frac{1}{\sqrt{x^3}})^3\cdot \frac{-\frac{3}{2}\cdot x^{\frac{1}{2}}}{x^3}-\frac{3\cdot 2x}{x^4}-5=-\frac{6}{x^7}-\frac{6}{x^3}-5$

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natsysan1 http://znanija.com/task/14840405

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