Срочно найдите производнкю функции. Created by rtyuioplkjhgf. algebra-ru

Срочно найдите производнкю функции...

армения20171

$y=\sqrt[3]{1+x\sqrt{x+3}}=\Big (1+x\sqrt{x+3}\Big )^{\frac{1}{3}}\\\\y'=\frac{1}{3}\cdot \Big (1+x\sqrt{x+3}\Big)^{-\frac{2}{3}}\cdot \Big (\sqrt{x+3}+x\cdot \frac{1}{2\sqrt{x+3}}\Big )=\\\\=\frac{1}{3\cdot \sqrt[3]{(1+x\sqrt{x+3})^2}}\cdot \frac{2(x+3)+x}{2\sqrt{x+3}}=\frac{1}{\sqrt[3]{(1+x\sqrt{x+3})^2}}\cdot \frac{x+1}{2\sqrt{x+3}}$

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армения20171 (2х+3)/2(√(х+3)) надо (3х+6)/(2√(х+3)

NNNLLL54

Verified answer

Y'=(³√(1+x√(x+3))'=1/3 *(1+x√(x+3))^(1/3-1)

*(√(x+3)+x*(√(x+3))'=

1/(3*(1+x√(x+3))^(2/3) *(√(x+3)+x *1/(2√(x+3))=
1/(3(1+x√(x+3))^(2/3) *
(2x+3)/(2√(x+3)

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армения20171 последний ( х+1) надо (х+2)

NNNLLL54 да

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