Final answer to the problem
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(8x+12\left(y^{\prime}x+y\right)+18y\cdot y^{\prime}\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
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Step-by-step Solution
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1
To derive the function $\frac{4x^2+12yx+9y^2}{3y+2x}$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\frac{4x^2+12yx+9y^2}{3y+2x}$
2
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\frac{4x^2+12yx+9y^2}{3y+2x}\right)$
Intermediate steps
3
Apply logarithm properties to both sides of the equality
$\ln\left(y\right)=\ln\left(4x^2+12yx+9y^2\right)-\ln\left(3y+2x\right)$
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4
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(\ln\left(4x^2+12yx+9y^2\right)-\ln\left(3y+2x\right)\right)$
5
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln\left(4x^2+12yx+9y^2\right)-\ln\left(3y+2x\right)\right)$
Intermediate steps
6
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(4x^2+12yx+9y^2\right)-\ln\left(3y+2x\right)\right)$
Explain this step further
7
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(4x^2+12yx+9y^2\right)\right)+\frac{d}{dx}\left(-\ln\left(3y+2x\right)\right)$
8
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(4x^2+12yx+9y^2\right)\right)-\frac{d}{dx}\left(\ln\left(3y+2x\right)\right)$
Intermediate steps
9
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\frac{d}{dx}\left(4x^2+12yx+9y^2\right)-\left(\frac{1}{3y+2x}\right)\frac{d}{dx}\left(3y+2x\right)$
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10
Multiplying the fraction by $-1$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\frac{d}{dx}\left(4x^2+12yx+9y^2\right)+\frac{-1}{3y+2x}\frac{d}{dx}\left(3y+2x\right)$
11
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(12yx\right)+\frac{d}{dx}\left(9y^2\right)\right)+\frac{-1}{3y+2x}\frac{d}{dx}\left(3y+2x\right)$
12
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(12yx\right)+\frac{d}{dx}\left(9y^2\right)\right)+\frac{-1}{3y+2x}\left(\frac{d}{dx}\left(3y\right)+\frac{d}{dx}\left(2x\right)\right)$
Intermediate steps
13
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(12yx\right)+\frac{d}{dx}\left(9y^2\right)\right)+\frac{-1}{3y+2x}\left(3\frac{d}{dx}\left(y\right)+\frac{d}{dx}\left(2x\right)\right)$
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Intermediate steps
14
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(12yx\right)+\frac{d}{dx}\left(9y^2\right)\right)+\frac{-1}{3y+2x}\left(3\frac{d}{dx}\left(y\right)+2\frac{d}{dx}\left(x\right)\right)$
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Intermediate steps
15
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(12yx\right)+\frac{d}{dx}\left(9y^2\right)\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\frac{d}{dx}\left(x\right)\right)$
Explain this step further
Intermediate steps
16
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(12yx\right)+\frac{d}{dx}\left(9y^2\right)\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
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Intermediate steps
17
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(4\frac{d}{dx}\left(x^2\right)+12\frac{d}{dx}\left(yx\right)+9\frac{d}{dx}\left(y^2\right)\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
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18
Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(4\frac{d}{dx}\left(x^2\right)+12\left(\frac{d}{dx}\left(y\right)x+y\frac{d}{dx}\left(x\right)\right)+9\frac{d}{dx}\left(y^2\right)\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
Intermediate steps
19
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(4\frac{d}{dx}\left(x^2\right)+12\left(y^{\prime}x+y\frac{d}{dx}\left(x\right)\right)+9\frac{d}{dx}\left(y^2\right)\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
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Intermediate steps
20
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(4\frac{d}{dx}\left(x^2\right)+12\left(y^{\prime}x+y\right)+9\frac{d}{dx}\left(y^2\right)\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
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21
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(4\frac{d}{dx}\left(x^2\right)+12\left(y^{\prime}x+y\right)+18y\frac{d}{dx}\left(y\right)\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
Intermediate steps
22
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(4\frac{d}{dx}\left(x^2\right)+12\left(y^{\prime}x+y\right)+18y\cdot y^{\prime}\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
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Intermediate steps
23
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(8x+12\left(y^{\prime}x+y\right)+18y\cdot y^{\prime}\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$
Explain this step further
Final answer to the problem$\frac{y^{\prime}}{y}=\frac{1}{4x^2+12yx+9y^2}\left(8x+12\left(y^{\prime}x+y\right)+18y\cdot y^{\prime}\right)+\frac{-1}{3y+2x}\left(3y^{\prime}+2\right)$