Ответ:
![y'=\frac{\cos(x)}{\sqrt{1+sin^2{x}}}](https://tex.z-dn.net/?f=y%27%3D%5Cfrac%7B%5Ccos%28x%29%7D%7B%5Csqrt%7B1%2Bsin%5E2%7Bx%7D%7D%7D)
Пошаговое объяснение:
![y=\ln(\sin(x)+\sqrt{1+sin^2(x)})](https://tex.z-dn.net/?f=y%3D%5Cln%28%5Csin%28x%29%2B%5Csqrt%7B1%2Bsin%5E2%28x%29%7D%29)
Сначала найдём производную натурального логарифма, затем производную подкоренного выражения
![y'=\frac{1}{\sin(x)+\sqrt{1+sin^2(x)}}\times(\cos(x)+\frac{1}{2\sqrt{1+sim^2(x)}}\times2\sin(x)\cos(x))=\\=\frac{1}{\sin(x)+\sqrt{1+sin^2(x)}}\times(\cos(x)+\frac{\sin(x)\cos(x)}{\sqrt{1+sim^2(x)}})=\\=\frac{1}{\sin(x)+\sqrt{1+sin^2(x)}}\times(\frac{\cos(x)\sqrt{1+\sin^2(x)}+\sin(x)\cos(x)}{\sqrt{1+sim^2(x)}})=\\=\frac{1}{\sin(x)+\sqrt{1+sin^2(x)}}\times(\frac{\cos(x)(\sqrt{1+\sin^2(x)}+\sin(x))}{\sqrt{1+sim^2(x)}})=\frac{\cos(x)}{\sqrt{1+sin^2{x}}}](https://tex.z-dn.net/?f=y%27%3D%5Cfrac%7B1%7D%7B%5Csin%28x%29%2B%5Csqrt%7B1%2Bsin%5E2%28x%29%7D%7D%5Ctimes%28%5Ccos%28x%29%2B%5Cfrac%7B1%7D%7B2%5Csqrt%7B1%2Bsim%5E2%28x%29%7D%7D%5Ctimes2%5Csin%28x%29%5Ccos%28x%29%29%3D%5C%5C%3D%5Cfrac%7B1%7D%7B%5Csin%28x%29%2B%5Csqrt%7B1%2Bsin%5E2%28x%29%7D%7D%5Ctimes%28%5Ccos%28x%29%2B%5Cfrac%7B%5Csin%28x%29%5Ccos%28x%29%7D%7B%5Csqrt%7B1%2Bsim%5E2%28x%29%7D%7D%29%3D%5C%5C%3D%5Cfrac%7B1%7D%7B%5Csin%28x%29%2B%5Csqrt%7B1%2Bsin%5E2%28x%29%7D%7D%5Ctimes%28%5Cfrac%7B%5Ccos%28x%29%5Csqrt%7B1%2B%5Csin%5E2%28x%29%7D%2B%5Csin%28x%29%5Ccos%28x%29%7D%7B%5Csqrt%7B1%2Bsim%5E2%28x%29%7D%7D%29%3D%5C%5C%3D%5Cfrac%7B1%7D%7B%5Csin%28x%29%2B%5Csqrt%7B1%2Bsin%5E2%28x%29%7D%7D%5Ctimes%28%5Cfrac%7B%5Ccos%28x%29%28%5Csqrt%7B1%2B%5Csin%5E2%28x%29%7D%2B%5Csin%28x%29%29%7D%7B%5Csqrt%7B1%2Bsim%5E2%28x%29%7D%7D%29%3D%5Cfrac%7B%5Ccos%28x%29%7D%7B%5Csqrt%7B1%2Bsin%5E2%7Bx%7D%7D%7D)
НОК (21, 28,35)= 3х7х2х2х5= 420
21 = 3х7
28=7х2х2
35=7х5
Периметр квадратной рамки будет 7*4=28(см), периметр треугольной рамки 9*3=27(см), не выйдет, он меньше, 1 см не хватит.
42 см3=0,042 дм3
............................