найдите производную ,пожалуйстаааа. Created by kristinakorostelova. algebra-ru

1) y' = (5x²+8x-6)' = (5x²)'+(8x)'+(6)' = 10x+8+0 = 10x+82) y' = [(x²+4x)(x-2)]' = (x²+4x)'(x-2)+(x²+4x)(x-2)' = (2x+4)(x-2)+(x²+4x) = = 2(x+2)(x-2)+x²+4x = 2(x²-4)+x²+4x = 2x²-8+x²+4x = 3x²+4x-83) 4) y' = (cos x²)' = -sin x² (x²)' = -2x sin x²5) y' = (sin²x)' = 2sin x (sin x)' = 2sin x cos x = sin2x

найдите производную ,пожалуйстаааа...

kristinakorostelova @kristinakorostelova

July 2022 1 10 Report

найдите производную ,пожалуйстаааа

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DeSex 1) y' = (5x²+8x-6)' = (5x²)'+(8x)'+(6)' = 10x+8+0 = 10x+8

2) y' = [(x²+4x)(x-2)]' = (x²+4x)'(x-2)+(x²+4x)(x-2)' = (2x+4)(x-2)+(x²+4x) =
= 2(x+2)(x-2)+x²+4x = 2(x²-4)+x²+4x = 2x²-8+x²+4x = 3x²+4x-8

3) $y' =( \frac{ x^{2} -4}{2x+1} )'= \frac{ (x^{2} -4)'(2x+1)-(x^{2} -4)(2x+1)'}{(2x+1)^2}=\\
= \frac{ 2x(2x+1)-2(x^{2} -4)}{(2x+1)^2}=\frac{ 4x^2+2x-2x^{2}+8}{(2x+1)^2}=
\frac{ 2x^2+2x+8}{(2x+1)^2}=\frac{ 2(x^2+x+4)}{(2x+1)^2}$

4) y' = (cos x²)' = -sin x² (x²)' = -2x sin x²

5) y' = (sin²x)' = 2sin x (sin x)' = 2sin x cos x = sin2x

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