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\frac{d}{dx}\left(\frac{-2x\cdot\left(x+4\right)}{\left(x^2-x-2\right)^2}\right)

Find the derivative of (x(x+4)*-2)/((-1x-2+x^2)^2)

Answer

$\frac{-8x\left(4+x\right)\left(\frac{d}{dx}\left(x^2\right)-1\right)-4x^2\left(4+x\right)\left(\frac{d}{dx}\left(x^2\right)-1\right)+4x^{3}\left(4+x\right)\left(\frac{d}{dx}\left(x^2\right)-1\right)+4\left(-2\left(4+x\right)-2x\right)+\left(4x-4x^2\right)\left(-2\left(4+x\right)-2x\right)+x^2\left(-2\left(4+x\right)-2x\right)-2x^{3}\left(-2\left(4+x\right)-2x\right)+x^{4}\left(-2\left(4+x\right)-2x\right)}{\left(-2-x+x^2\right)^{4}}$

Step-by-step explanation

Problem

$\frac{d}{dx}\left(\frac{-2x\cdot\left(x+4\right)}{\left(x^2-x-2\right)^2}\right)$
1

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{2x\left(4+x\right)\frac{d}{dx}\left(\left(-2-x+x^2\right)^2\right)+\left(-2-x+x^2\right)^2\frac{d}{dx}\left(-2x\left(4+x\right)\right)}{\left(\left(-2-x+x^2\right)^2\right)^2}$

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Answer

$\frac{-8x\left(4+x\right)\left(\frac{d}{dx}\left(x^2\right)-1\right)-4x^2\left(4+x\right)\left(\frac{d}{dx}\left(x^2\right)-1\right)+4x^{3}\left(4+x\right)\left(\frac{d}{dx}\left(x^2\right)-1\right)+4\left(-2\left(4+x\right)-2x\right)+\left(4x-4x^2\right)\left(-2\left(4+x\right)-2x\right)+x^2\left(-2\left(4+x\right)-2x\right)-2x^{3}\left(-2\left(4+x\right)-2x\right)+x^{4}\left(-2\left(4+x\right)-2x\right)}{\left(-2-x+x^2\right)^{4}}$
$\frac{d}{dx}\left(\frac{-2x\cdot\left(x+4\right)}{\left(x^2-x-2\right)^2}\right)$

Main topic:

Differential calculus

Used formulas:

5. See formulas

Time to solve it:

~ 12.11 seconds