Question

найдите значение производной функции в точке x0
Помогите пожалуйста.
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Answer 1

Ответ:

$1)\ \ f(x)=3ctgx+2sinx\\\\f'(x)=-\dfrac{3}{sin^2x}+2\, cosx\\\\f'(\dfrac{\pi}{6})=-\dfrac{3}{1/4}+2\cdot \dfrac{\sqrt3}{2}=-12+\sqrt3\\\\\\2)\ \ f(x)=x-2cosx+3tgx\\\\f'(x)=1+2sinx+\dfrac{3}{cos^2x}\\\\f'(\dfrac{\pi}{4})=1+2\cdot \dfrac{\sqrt2}{2}+\dfrac{3}{1/2}=1+\sqrt2+6=7+\sqrt2\\\\\\3)\ \ f(x)=2x^2-2tgx+sinx\\\\f'(x)=4x-\dfrac{2}{cos^2x}+cosx\\\\f'(\dfrac{3\pi}{4})=3\pi-\dfrac{2}{1/2}-\dfrac{\sqrt2}{2}=3\pi -4-\dfrac{\sqrt2}{2}$

$4)\ \ f(x)=\dfrac{2}{x}-2ctgx+4sinx\\\\f'(x)=-\dfrac{2}{x^2}+\dfrac{2}{sin^2x}+4cosx\\\\f'(\dfrac{2\pi}{3})=-\dfrac{2}{4\pi ^2/9}+\dfrac{2}{3/4}-\dfrac{4}{2}=-\dfrac{9}{2\pi ^2}+\dfrac{8}{3}-2=-\dfrac{9}{2\pi ^2}+\dfrac{2}{3}$