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Step-by-step Solution

Derive the function $\ln\left(2x^2+4x\right)$ with respect to x

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Answer

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

Step-by-step explanation

Problem to solve:

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

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 $\displaystyle f'(x)=\frac{a'}{a}$

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

The derivative of a sum of two functions is the sum of the derivatives of each function

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

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Answer

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

Main topic:

Differential calculus

Used formulas:

7. See formulas

Time to solve it:

~ 0.82 seconds

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