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Step-by-step Solution

Find the derivative using the quotient rule (d/dx)((ln(x^2+4*x)/(x^2+4x))

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Answer

$\frac{-4x^2\ln\left(x^2+4x\right)-16x\ln\left(x^2+4x\right)-2x^{3}\ln\left(x^2+4x\right)-8x^2\ln\left(x^2+4x\right)+2x^2x+16x+12x^2}{\left(x^2+4x\right)^{3}}$

Step-by-step explanation

Problem to solve:

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

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

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

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

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Answer

$\frac{-4x^2\ln\left(x^2+4x\right)-16x\ln\left(x^2+4x\right)-2x^{3}\ln\left(x^2+4x\right)-8x^2\ln\left(x^2+4x\right)+2x^2x+16x+12x^2}{\left(x^2+4x\right)^{3}}$
$\frac{d}{dx}\left(\frac{\ln\left(x^2+4x\right)}{x^2+4x}\right)$

Main topic:

Differential calculus

Used formulas:

6. See formulas

Time to solve it:

~ 1.68 seconds

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