Final answer to the problem
$\frac{\left(-4x-2x^2-2\right)\left(1-x\right)}{\left(x+1\right)^{2}}$
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Step-by-step Solution
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1
To derive the function $\frac{\left(1-x^2\right)^2}{x^2+2x+1}$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
2
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\frac{\left(1-x^2\right)^2}{x^2+2x+1}\right)$
Intermediate steps
3
Apply logarithm properties to both sides of the equality
$\ln\left(y\right)=2\ln\left(1-x^2\right)-\ln\left(x^2+2x+1\right)$
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4
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(2\ln\left(1-x^2\right)-\ln\left(x^2+2x+1\right)\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(2\ln\left(1-x^2\right)-\ln\left(x^2+2x+1\right)\right)$
Intermediate steps
6
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(2\ln\left(1-x^2\right)-\ln\left(x^2+2x+1\right)\right)$
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7
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(2\ln\left(1-x^2\right)\right)+\frac{d}{dx}\left(-\ln\left(x^2+2x+1\right)\right)$
Intermediate steps
8
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=2\frac{d}{dx}\left(\ln\left(1-x^2\right)\right)-\frac{d}{dx}\left(\ln\left(x^2+2x+1\right)\right)$
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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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(1-x^2\right)-\left(\frac{1}{x^2+2x+1}\right)\frac{d}{dx}\left(x^2+2x+1\right)$
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10
Multiplying the fraction by $-1$
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(1-x^2\right)+\frac{-1}{x^2+2x+1}\frac{d}{dx}\left(x^2+2x+1\right)$
11
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\left(\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(-x^2\right)\right)+\frac{-1}{x^2+2x+1}\frac{d}{dx}\left(x^2+2x+1\right)$
12
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\left(\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(-x^2\right)\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right)\right)$
13
The derivative of the constant function ($1$) is equal to zero
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right)\right)$
14
The derivative of the constant function ($1$) is equal to zero
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)$
Intermediate steps
15
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(x\right)\right)$
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Intermediate steps
16
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+2\right)$
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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
$\frac{y^{\prime}}{y}=-2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+2\right)$
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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 $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=-4\left(\frac{1}{1-x^2}\right)x+\frac{-1}{x^2+2x+1}\left(2x+2\right)$
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19
Multiply the fraction and term
$\frac{y^{\prime}}{y}=\frac{-4x}{1-x^2}+\frac{-1}{x^2+2x+1}\left(2x+2\right)$
20
Multiplying the fraction by $2x+2$
$\frac{y^{\prime}}{y}=\frac{-4x}{1-x^2}+\frac{-\left(2x+2\right)}{x^2+2x+1}$
21
Simplify the product $-(2x+2)$
$\frac{y^{\prime}}{y}=\frac{-4x}{1-x^2}+\frac{-2x-2}{x^2+2x+1}$
22
Multiply both sides of the equation by $y$
$y^{\prime}=\left(\frac{-4x}{1-x^2}+\frac{-2x-2}{x^2+2x+1}\right)y$
23
Substitute $y$ for the original function: $\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
$y^{\prime}=\left(\frac{-4x}{1-x^2}+\frac{-2x-2}{x^2+2x+1}\right)\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
24
The derivative of the function results in
$\left(\frac{-4x}{1-x^2}+\frac{-2x-2}{x^2+2x+1}\right)\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
Intermediate steps
25
Simplify the derivative
$\frac{\left(-4x-2x^2-2\right)\left(1-x\right)}{\left(x+1\right)^{2}}$
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Final answer to the problem$\frac{\left(-4x-2x^2-2\right)\left(1-x\right)}{\left(x+1\right)^{2}}$