Задача 65032 Найти производную [m]y = arcsin...

60b77ffca1c50e1d3043a110

10.06.2022 12:14:49

Найти производную [m]y = arcsin \frac{sin x}{\sqrt{1+sin^2 x}}[/m]

626fab9a354ab65fb2f43abb

10.06.2022 12:21:19

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[m]y = arcsin \frac{sin x}{\sqrt{1+sin^2 x}}[/m]
[m]y ' = \frac{1}{\sqrt{1-(\frac{sin x}{\sqrt{1+sin^2 x}})^2}}*\frac{cos x*\sqrt{1+sin^2 x} - sin x*2sin x*cos x/(2\sqrt{1+sin^2 x})}{1+sin^2 x} =[/m]
[m]= \frac{1}{\sqrt{1-\frac{sin^2 x}{1+sin^2 x}}}*\frac{cos x*\sqrt{1+sin^2 x} - sin^2 x*cos x/\sqrt{1+sin^2 x}}{1+sin^2 x} =[/m]
[m]= \frac{1}{\sqrt{\frac{1+sin^2 x - sin^2 x}{1+sin^2 x}}}*\frac{cos x*\sqrt{1+sin^2 x} - sin^2 x*cos x/\sqrt{1+sin^2 x}}{1+sin^2 x} =[/m]
[m]= \frac{sqrt{1+sin^2 x}}{\sqrt{1}}*\frac{cos x*\sqrt{1+sin^2 x} - sin^2 x*cos x/\sqrt{1+sin^2 x}}{1+sin^2 x} =[/m]
[m]= \frac{cos x*\sqrt{1+sin^2 x} - sin^2 x*cos x/\sqrt{1+sin^2 x}}{\sqrt{1+sin^2 x}} =[/m]
[m]= \frac{cos x(1+sin^2 x) - sin^2 x*cos x}{\sqrt{1+sin^2 x}*\sqrt{1+sin^2 x}} =[/m]
[m]= \frac{cos x(1+sin^2 x) - sin^2 x*cos x}{1+sin^2 x} =[/m]