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

Derive the function (ln(x^2+4x))/(x^2+4x) with respect to x

Answer

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

Step-by-step explanation

Problem

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

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

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Answer

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

Main topic:

Differential calculus

Used formulas:

4. See formulas

Time to solve it:

~ 0.39 seconds