Найдите производную второго порядка от y=[tex]\sqrt{1+x^{2}[/tex]. Created by AndrewShabalin. algebra-ru

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July 2022 1 4 Report

Найдите производную второго порядка от y=[tex]\sqrt{1+x^{2}[/tex]

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natalyabryukhova

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Ответ:

$y=\sqrt{1+x^2}\\y'=\frac{1}{2}*(1+x^2)^-^\frac{1}{2}*2x=\frac{x}{\sqrt{1+x^2}}\\y''=\frac{\sqrt{1+x^2}-x*(1+x^2)^-^\frac{1}{2}*2x}{1+x^2} =\frac{1+x^2-2x^2}{\sqrt{1+x^2}(1+x^2)}= \frac{1-x^2}{(1+x^2)\sqrt{1+x^2} }$

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tex], y=0...

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tex], y=x+1...

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najdite uravnenie ploskosti prohodyashej cherez tochku peresecheniya ploskostej 2x2y

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diagonal bokovoj grani pravilnoj treugolnoj prizmy ravna43 i sostavlyaet ugo

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Answers & Comments

Verified answer

najdite obratnuyu matricu dlya 3 4 5 7

dan sleduyushij variacionnyj ryad 1 2 3 4 5 6 7 8 9 10 1 1 2 2 4 4 4 5 5 5 trebuets

tex], y=0...

tex], y=x+1...

najdite uravnenie ploskosti prohodyashej cherez tochku peresecheniya ploskostej 2x2y

diagonal bokovoj grani pravilnoj treugolnoj prizmy ravna43 i sostavlyaet ugo

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