Помогите пожалуйста!. Created by polin252. algebra-ru

Помогите пожалуйста!...

polin252 @polin252

November 2021 1 7 Report

Помогите пожалуйста!

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NNNLLL54

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$1)\ \ f(x)=-3x^5\ \ ,\ \ f'(x)=-3\cdot 5x^4=-15x^4\\\\f(x)=-5x^{-8}\ \ ,\ \ f'(x)=-5\cdot (-8)x^{-9}=40x^{-9}\\\\f(x)=\dfrac{4}{x^5}\ \ ,\ \ f'(x)=-\dfrac{4\cdot 5x^4}{x^{10}}=-\dfrac{20}{x^6}\\\\f(x)=x^{\frac{7}{6}}\ \ ,\ \ f'(x)=\dfrac{7}{6}\cdot x^{\frac{1}{6}}\\\\f(x)=(3x-7)(1-2x)=17x-6x^2-7\ \ ,\ \ \ f'(x)=17-12x\\\\f(x)=\dfrac{7-10x}{5x+2}\ \ ,\ \ f'(x)=\dfrac{-10(5x+2)-5(7-10x)}{(5x+2)^2}=-\dfrac{55}{(5x+2)^2}$

$2)\ \ f(x)=3x^2-x^3\ \ ,\ \ x_0=-2\ \ ,\ \ \ \ \ y=f(x_0)+f'(x_0)(x-x_0)\\\\f(-2)=3\cdot 4-8=12-8=4\\\\f'(x)=6x-3x^2\ \ ,\ \ f'(-2)=-12-3\cdot 4=-12-12=-24\\\\y=4-24\, (x+2)\\\\y=-24x-44$

$3)\ \ f(x)=\dfrac{cosx}{1-x}\ \ ,\ \ \ x_0=0\\\\\\f'(x)=\dfrac{-sinx\, (1-x)-cosx\cdot (-1)}{(1-x)^2}=\dfrac{-sinx+x\cdot sinx+cosx}{(1-x)^2}\\\\\\f'(0)=\dfrac{-sin0-0\cdot sin0+cos0}{(1-0)^2}=\dfrac{1}{1}=1$

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polin252 Спасибо большое

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