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\frac{d}{dx}\left(\frac{2}{3}\cdot arccos\left(\sqrt{x^3+3x}\right)\right)

Derive the function 2/3arccos((x^3+3x)^0.5) with respect to x

Answer

$\frac{-x^{2}-1}{3x\sqrt{1-3x-x^3}\sqrt{3x+x^3}+x^3\sqrt{1-3x-x^3}\sqrt{3x+x^3}}$

Step-by-step explanation

Problem

$\frac{d}{dx}\left(\frac{2}{3}\cdot arccos\left(\sqrt{x^3+3x}\right)\right)$
1

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$\frac{2}{3}\cdot\frac{d}{dx}\left(arccos\left(\sqrt{3x+x^3}\right)\right)$

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Answer

$\frac{-x^{2}-1}{3x\sqrt{1-3x-x^3}\sqrt{3x+x^3}+x^3\sqrt{1-3x-x^3}\sqrt{3x+x^3}}$

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$\frac{d}{dx}\left(\frac{2}{3}\cdot arccos\left(\sqrt{x^3+3x}\right)\right)$

Main topic:

Differential calculus

Used formulas:

4. See formulas

Time to solve it:

~ 0.49 seconds