Final Answer
$\frac{8+4x}{\left(x+3\right)^2\left(x+2\right)}$
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Step-by-step Solution
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1
To derive the function $\frac{x^2+x-2}{x^2+5x+6}$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\frac{x^2+x-2}{x^2+5x+6}$
2
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\frac{x^2+x-2}{x^2+5x+6}\right)$
Intermediate steps
3
Apply logarithm properties to both sides of the equality
$\ln\left(y\right)=\ln\left(x^2+x-2\right)-\ln\left(x^2+5x+6\right)$
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4
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(\ln\left(x^2+x-2\right)-\ln\left(x^2+5x+6\right)\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln\left(x^2+x-2\right)-\ln\left(x^2+5x+6\right)\right)$
Intermediate steps
6
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(x^2+x-2\right)-\ln\left(x^2+5x+6\right)\right)$
Explain this step further
7
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(x^2+x-2\right)\right)+\frac{d}{dx}\left(-\ln\left(x^2+5x+6\right)\right)$
8
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(x^2+x-2\right)\right)-\frac{d}{dx}\left(\ln\left(x^2+5x+6\right)\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\frac{d}{dx}\left(x^2+x-2\right)-\left(\frac{1}{x^2+5x+6}\right)\frac{d}{dx}\left(x^2+5x+6\right)$
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10
Multiplying the fraction by $-1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\frac{d}{dx}\left(x^2+x-2\right)+\frac{-1}{x^2+5x+6}\frac{d}{dx}\left(x^2+5x+6\right)$
11
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-2\right)\right)+\frac{-1}{x^2+5x+6}\frac{d}{dx}\left(x^2+5x+6\right)$
12
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-2\right)\right)+\frac{-1}{x^2+5x+6}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(5x\right)+\frac{d}{dx}\left(6\right)\right)$
13
The derivative of the constant function ($-2$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)\right)+\frac{-1}{x^2+5x+6}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(5x\right)+\frac{d}{dx}\left(6\right)\right)$
14
The derivative of the constant function ($6$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)\right)+\frac{-1}{x^2+5x+6}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(5x\right)\right)$
Intermediate steps
15
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(\frac{d}{dx}\left(x^2\right)+1\right)+\frac{-1}{x^2+5x+6}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(5x\right)\right)$
Explain this step further
Intermediate steps
16
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(\frac{d}{dx}\left(x^2\right)+1\right)+\frac{-1}{x^2+5x+6}\left(\frac{d}{dx}\left(x^2\right)+5\frac{d}{dx}\left(x\right)\right)$
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Intermediate steps
17
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(\frac{d}{dx}\left(x^2\right)+1\right)+\frac{-1}{x^2+5x+6}\left(\frac{d}{dx}\left(x^2\right)+5\right)$
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Intermediate steps
18
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=\frac{1}{x^2+x-2}\left(2x+1\right)+\frac{-1}{x^2+5x+6}\left(2x+5\right)$
Explain this step further
19
Multiply the fraction and term
$\frac{y^{\prime}}{y}=\frac{2x+1}{x^2+x-2}+\frac{-1}{x^2+5x+6}\left(2x+5\right)$
20
Multiplying the fraction by $2x+5$
$\frac{y^{\prime}}{y}=\frac{2x+1}{x^2+x-2}+\frac{-\left(2x+5\right)}{x^2+5x+6}$
21
Factor the trinomial $x^2+x-2$ finding two numbers that multiply to form $-2$ and added form $1$
$\begin{matrix}\left(-1\right)\left(2\right)=-2\\ \left(-1\right)+\left(2\right)=1\end{matrix}$
$\frac{y^{\prime}}{y}=\frac{2x+1}{\left(x-1\right)\left(x+2\right)}+\frac{-\left(2x+5\right)}{x^2+5x+6}$
23
Factor the trinomial $x^2+5x+6$ finding two numbers that multiply to form $6$ and added form $5$
$\begin{matrix}\left(2\right)\left(3\right)=6\\ \left(2\right)+\left(3\right)=5\end{matrix}$
$\frac{y^{\prime}}{y}=\frac{2x+1}{\left(x-1\right)\left(x+2\right)}+\frac{-\left(2x+5\right)}{\left(x+2\right)\left(x+3\right)}$
25
Simplify the product $-(2x+5)$
$\frac{y^{\prime}}{y}=\frac{2x+1}{\left(x-1\right)\left(x+2\right)}+\frac{-2x-5}{\left(x+2\right)\left(x+3\right)}$
26
Multiply both sides of the equation by $y$
$y^{\prime}=\left(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}+\frac{-2x-5}{\left(x+2\right)\left(x+3\right)}\right)y$
27
Substitute $y$ for the original function: $\frac{x^2+x-2}{x^2+5x+6}$
$y^{\prime}=\left(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}+\frac{-2x-5}{\left(x+2\right)\left(x+3\right)}\right)\frac{x^2+x-2}{x^2+5x+6}$
28
The derivative of the function results in
$\left(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}+\frac{-2x-5}{\left(x+2\right)\left(x+3\right)}\right)\frac{x^2+x-2}{x^2+5x+6}$
Intermediate steps
29
Simplify the derivative
$\frac{8+4x}{\left(x+3\right)^2\left(x+2\right)}$
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Final Answer$\frac{8+4x}{\left(x+3\right)^2\left(x+2\right)}$