Question

Допоможіть 100б..........

Answer 1

№8

Для функции   $f(x) = \sin 4x + \cos 4x$   найдите  $f'\bigg(\dfrac{\pi }{8} \bigg)$

$\bullet ~ \Big(f(g(x)) \Big) ' = f'(g(x))\cdot g'(x) \\\\ \bullet ~~(u \pm v )' = u'\pm v'$

                                                 

$f'(x) = (\sin 4x + \cos 4x) ' = \cos 4x\cdot (4x)'- \sin 4x \cdot (4x)' = \\\\= 4(\cos 4x - \sin 4x)$

$f'\bigg(\dfrac{\pi }{8} \bigg) = 4(~ \cos (4\cdot \frac{\pi }{8}) - \sin (4\cdot \frac{\pi }{8}) ~) = 4\cdot (\cos 90 - \sin 90) = 4(0-1) = - 4$

№9

Найдите значение производной функции

$y = \cos 2x \cdot tg(x-\frac{\pi }{4} )$ в точке  $x_0 =\frac{\pi }{2}$

$\bullet ~ \Big(f(g(x)) \Big) ' = f'(g(x))\cdot g'(x) \\\\ \bullet ~~(u \cdot v )' = u'v + uv'$

$y' =( \cos 2x \cdot tg(x-\frac{\pi }{4} ) ~)' = (\cos 2x)' \cdot tg (x-\frac{\pi }{4} )+ \cos 2x \cdot (tg (x-\frac{\pi }{4} )~) ' = \\\\\\ =- 2 \sin 2x\cdot tg (x-\frac{\pi }{4} ) + \cos 2x \cdot \dfrac{1}{\cos^2(x-\frac{\pi }{4} )}$

$y'(\frac{\pi }{2} ) \displaystyle = - 2 \cdot \sin \pi \cdot tg \tfrac{\pi }{4}+ \cos \pi \cdot \frac{1}{\cos ^2\frac{\pi }{4} } = -2 \cdot 0 \cdot 1 + (-1)\cdot \frac{1}{\dfrac{1}{2} } =\\\\= 0 -2 = -2$