Question

Найдите пожалуйста f’(x)!!! Задание на фото.
(50 баллов)











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Answer 1

Ответ.

   $\boxed{\ (x^{n})'=n\, x^{n-1}\ \ ,\ \ (Cu)'=C\cdot u'\ \ ,\ \ x'=1\ \ ,\ \ C'=0\ }$

$1)\ \ f(x)=\dfrac{1}{3}x^3-x^2+2x\ \ ,\ \ f'(x)=\dfrac{1}{3}\cdot 3x^2-2x+2=x^2-2x+2\\\\\\2)\ \ f(x)=\dfrac{2}{x^3}-x\sqrt{x}}=2x^{-3}-x^{\frac{3}{2}}\\\\f'(x)=-6x^{-4}-\dfrac{3}{2}\, x^{\frac{1}{2}}=-\dfrac{6}{x^4}-\dfrac{3}{2}\, \sqrt{x}\\\\\\3)\ \ f(x)=4sinx\cdot cosx=2sin2x\\\\f'(x)=2\cdot cos2x\cdot 2=4\, cos2x\ \ ,\\\\f'(-\frac{2\pi}{3})=4\cdot cos(-\frac{2\pi}{3})=4\cdot cos\frac{2\pi}{3}=4\cdot cos(\pi -\frac{\pi}{3})=-4\cdot cos\dfrac{\pi}{3}=-4\cdot \dfrac{1}{2}=-2$

$4)\ \ f(x)=\dfrac{2-3x}{x+2}\ \ ,\ \ \ \ \ \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}\\\\\\f'(x)=\dfrac{-3(x+2)-(2-3x)\cdot 1}{(x+2)^2}=\dfrac{-8}{(x+2)^2}\\\\f'(-1)=-\dfrac{8}{(-1+2)^2}=-8\\\\\\5)\ \ f(x)=\dfrac{1}{3}\, x^3-4x\ \ ,\ \ \ f'(x)=x^2-4=0\\\\(x-2)(x+2)=0\\\\x_1=-2\ ,\ x_2=2$