Step-by-step Solution

Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{-2x\left(x+4\right)}{\left(x^2-x-2\right)^2}\right)$

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

Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{-2x\cdot\left(x+4\right)}{\left(x^2-x-2\right)^2}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

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

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

Quotient rule

~ 0.92 seconds

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