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# Step-by-step Solution

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## Answer

$\frac{-32x}{\left(x^2+4\right)^2}$

## Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{16}{x^2+4}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\left(x^2+4\right)\frac{d}{dx}\left(16\right)-16\frac{d}{dx}\left(x^2+4\right)}{\left(x^2+4\right)^2}$
2

The derivative of the constant function is equal to zero

$\frac{0\left(x^2+4\right)-16\frac{d}{dx}\left(x^2+4\right)}{\left(x^2+4\right)^2}$

## Answer

$\frac{-32x}{\left(x^2+4\right)^2}$
$\frac{d}{dx}\left(\frac{16}{x^2+4}\right)$

### Main topic:

Differential calculus

~ 1.2 seconds

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