Вычислить производную в точке. 1) F(x)=3x^4 - 2x^2 + 4x-1 ; x= -1 2) F(x)= (3-x^2)*(4+x^2) ; в.т x= -2 3) F(x)= x^2+1/x^3+1 в.т x=1. Created by yasha11111. matematika-ru

x^3+1 в.т x=1...

July 2022 1 12 Report

Вычислить производную в точке.
1) F(x)=3x^4 - 2x^2 + 4x-1 ; x= -1
2) F(x)= (3-x^2)*(4+x^2) ; в.т x= -2
3) F(x)= x^2+1/x^3+1 в.т x=1

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$f(x)=3x^4-2x^2+4x-1\\f`(x)=3*4x^3-2*2x+4=12x^3-4x+4\\f`(-1)=12(-1)^3-4(-1)+4=-12+4+4=-4\\\\f(x)=(3-x^2)(4+x^2)\\f`(x)=(3-x^2)`(4+x^2)+(3-x^2)(4+x^2)`=\\=-2x(4+x^2)+2x(3-x^2)=-8x-2x^2+6x-2x^2=-4x^2-2x\\f`(-2)=-4(-2)^2-2(-2)=-4*2+4=-8+4=-4\\\\f(x)= \frac{x^2+1}{x^3+1}\\\\f`(x)= \frac{(x^2+1)`(x^3+1)-(x^2+1)(x^3+1)`}{(x^3+1)^2}= \frac{2x(x^3+1)-3x^2(x^2+1)}{(x^3+1)^2}=\\\\= \frac{2x^4+2x-3x^4-3x^2}{(x^3+1)^2}= \frac{-x^4-3x^2+2x}{(x^3+1)^2}$

$f`(1)= \frac{-1^4-3*1^2+2*1}{(1^3+1)^2}= \frac{-1-3+2}{(1+1)^2}= \frac{-2}{2^2}= \frac{-2}{4}=-0,5$

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