Question

Помогите решить ..............................

Answer 1

$17.14.\ \ a)\ \ f(x)=\sqrt{\dfrac{\sqrt3}{2}-cosx}\ \ \ ,\ \ \ \ \ \ \star \ \ (\sqrt{u})'=\dfrac{1}{2\sqrt{u}}\cdot u'\ \ \star \\\\\\f'(x)=\dfrac{1}{2\sqrt{\dfrac{\sqrt3}{2}-cosx}}\cdot sinx=\dfrac{\sqrt2\, sinx}{2\sqrt{\sqrt3-2cosx}}\\\\\\OOF\ \ f'(x):\ \ \sqrt3-2cosx>0\ \ ,\ \ \ cosx<\dfrac{\sqrt3}{2}\\\\\\\dfrac{\pi}{3}+2\pi n<x<\dfrac{5\pi}{3}+2\pi n\ ,\ n\in Z\\\\\\x\in D(y)=\Big(\dfrac{\pi}{3}+2\pi n\ ;\ \dfrac{5\pi}{3}+2\pi n\ \Big)\ ,\ n\in Z$

$b)\ \ f(x)=\sqrt{\dfrac{1}{2}+sinx}\\\\\\f'(x)=\dfrac{1}{2\sqrt{\dfrac{1}{2}+sinx}}\cdot cosx=\dfrac{\sqrt2\, cosx}{2\sqrt{1+2sinx}}\\\\\\OOF\ f'(x):\ \ \ 1+2sinx>0\ \ ,\ \ sinx<-\dfrac{1}{2}\\\\\\\dfrac{7\pi }{6}+2\pi n<x<\dfrac{11\pi }{6}+2\pi n\ ,\ n\in Z\\\\\\x\in D(y)=\Big(\dfrac{7\pi }{6}+2\pi n\ ;\ \dfrac{11\pi }{6}+2\pi n\, \Big)\ ,\ n\in Z$

$17.15.\ \ a)\ \ f(x)=sin(x^2+x+\frac{\pi }{4})\\\\f'(x)=cos(x^2+x+\frac{\pi}{4})\cdot (2x+1)\\\\x_0=0:\ \ \ f'(0)=cos\dfrac{\pi}{4}\cdot 1=\dfrac{\sqrt2}{2}\\\\\\b)\ \f(x)=\dfrac{4}{3}\cdot tg(x^3+x)\\\\\\f'(x)=\dfrac{4}{3}\cdot \dfrac{1}{cos^2(x^3+x)}\cdot (3x^2+1)\\\\\\x_0=0:\ \ \ f'(0)=\dfrac{4\cdot (0+1)}{3\cdot cos^20}=\dfrac{4}{3\cdot 1}=\dfrac{4}{3}$