Ответ(ы) на вопрос:
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1.
[latex]y= \frac{lnx}{sinx}+x*ctgx \\ y'= \frac{ \frac{sinx}{x}-lnx*cosx}{sin^2x}+ctgx- \frac{x}{sin^2x} = \frac{sinx-x*lnx*cosx}{x*sin^2x}+ \frac{cosx}{sinx}- \frac{x}{sin^2x}= [/latex][latex] \frac{sinx-x*lnx*cosx+x*sinx*cosx-x^2}{xsin^2x} =\frac{sinx-x^2+x*cosx(sinx-lnx)}{xsin^2x}[/latex]
2.
[latex]y= \frac{ \sqrt{x}}{\sqrt{x}+1} \\y'= \frac{ \frac{ \sqrt{x}+1}{2\sqrt{x}}-\frac{ \sqrt{x}}{2\sqrt{x}}}{(\sqrt{x}+1)^2}= \frac{1}{2\sqrt{x}*(\sqrt{x}+1)^2}[/latex]
3.
[latex] y=\frac{ctgx}{ \sqrt{x}} \\ y'= \frac{- \frac{ \sqrt{x}}{sin^2x}- \frac{ctgx}{2 \sqrt{x} } }{x}=- \frac{2x+sinx*cosx}{2x \sqrt{x} *sin^2x} =- \frac{4x+2sinx*cosx}{4x \sqrt{x} *sin^2x} =- \frac{4x+sin2x}{4x \sqrt{x} *sin^2x} [/latex]
4.
y = ln (x² + 2x)
[latex]y'= \frac{2x+2}{x^2+2x}= \frac{2(x+1)}{x(x+2)} [/latex]
5.
[latex]y=ln \frac{x^2}{1-x^2} \\ y'=\frac{1-x^2}{x^2}*\frac{2x(1-x^2)+x^2*2x}{(1-x^2)^2}=\frac{2}{x(1-x^2)}[/latex]
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