Question

Y=cos2x/sin(x+pi/4) {0;pi/2} производная

Answer 1

Находим производную функции. Воспользуемся формулой производной частного.
$y^{'}=...= \dfrac{(\cos 2x)^{'}\cdot \sin(x+ \frac{\pi}{4}) -\cos 2x\cdot (\sin (x+\frac{\pi}{4}))^{'}}{\sin^2(x+\frac{\pi}{4})} =\\\\ = \dfrac{-2\sin 2x\cdot \sin(x+\frac{\pi}{4})-\cos 2x\cdot \cos(x+\frac{\pi}{4})}{\sin^2(x+\frac{\pi}{4})}$
Приравниваем производную функции к нулю:
$-2\sin 2x\cdot\sin(x+\frac{\pi}{4})-\cos 2x\cdot\cos (x+\frac{\pi}{4})=0|:\cos 2x\cos(x+\frac{\pi}{4})\\ -2tg 2x\cdot tg(x+\frac{\pi}{4})-1=0$
Воспользуемся формулами:
  $tg (a+b)= \dfrac{tga + tg b}{1-tg a\cdot tgb}$ и  $tg 2x= \dfrac{2tg x}{1-tg^2x}$

$-2\cdot \dfrac{2tg x}{1-tg^2x} \cdot \dfrac{tgx+1}{1-tg x} -1=0 \\ \\ - \dfrac{4tg x}{(1-tg x)(1+tg x)} \cdot \dfrac{tg x+1}{1-tg x} -1=0\\ \\ \dfrac{-4tg x}{(1-tg x)^2} -1=0$
Приводим дробь к общему знаменателю:
$\dfrac{-4tg x-(1-tgx)^2}{(1-tg x)^2} =0$
Дробь обращается в нуль, если числитель равен нулю :
$-4tg x-(1-tgx)^2=0\\ -4tg x-1+2tgx-tg^2x=0\\ -tg^2x-2tg x-1=0\\ -(tg^2x+2tg x+1)=0\\ -(tg x+1)^2=0\\ tg x = -1\\ \\ x=- \dfrac{\pi}{4}+ \pi n ,n \in \mathbb{Z}$

Подберем корни, которые принадлежат заданному отрезку:
$n=1;\,\,\,\,\, x=- \dfrac{\pi}{4}+\pi = \dfrac{3\pi}{4}$ - не принадлежит заданному отрезку.

Вычислим значение функции на отрезке:
$y(0)= \dfrac{\cos(2\cdot 0)}{\sin (0+ \dfrac{\pi}{4} )} = \dfrac{1}{ \dfrac{1}{\sqrt{2}} } =\sqrt{2}$ - наибольшее значение


$y( \dfrac{\pi}{2} )= \dfrac{\cos(2\cdot \dfrac{\pi}{2}) }{\sin ( \dfrac{\pi}{2} + \dfrac{\pi}{4} )} = \dfrac{-1}{ \dfrac{1}{\sqrt{2}} } =-\sqrt{2}$ - наименьшее значение

Answer 2

To find the derivative of the function Y = (cos(2x)) / sin(x + π/4) over the interval [0, π/2], we can apply the quotient rule. Let's go step by step.

The quotient rule states that for two functions u(x) and v(x), the derivative of their quotient u(x) / v(x) is given by:

(d/dx)[u(x) / v(x)] = [v(x) * u'(x) - u(x) * v'(x)] / [v(x)]^2

In our case, u(x) = cos(2x) and v(x) = sin(x + π/4). Let's calculate the derivatives:

u'(x) = d/dx [cos(2x)] = -sin(2x) * 2 = -2sin(2x)

v'(x) = d/dx [sin(x + π/4)] = cos(x + π/4)

Now, substitute these derivatives into the quotient rule formula:

Y' = [(sin(x + π/4)) * (-2sin(2x)) - (cos(2x)) * (cos(x + π/4))] / [(sin(x + π/4))^2]

Simplifying further:

Y' = [-2sin(x + π/4)sin(2x) - cos(2x)cos(x + π/4)] / [sin^2(x + π/4)]

Finally, we have the derivative of the function Y = (cos(2x)) / sin(x + π/4) as:

Y' = [-2sin(x + π/4)sin(2x) - cos(2x)cos(x + π/4)] / [sin^2(x + π/4)]