Exercise$\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
Step-by-step Solution
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
$\ln\left(y\right)=2\ln\left(1-x^2\right)-\ln\left(x^2+2x+1\right)$
Explain this step further
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(2\ln\left(1-x^2\right)-\ln\left(x^2+2x+1\right)\right)$
6
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(2\ln\left(1-x^2\right)-\ln\left(x^2+2x+1\right)\right)$
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
$\frac{y^{\prime}}{y}=2\frac{d}{dx}\left(\ln\left(1-x^2\right)\right)-\frac{d}{dx}\left(\ln\left(x^2+2x+1\right)\right)$
Explain this step further
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(1-x^2\right)-\left(\frac{1}{x^2+2x+1}\right)\frac{d}{dx}\left(x^2+2x+1\right)$
Explain this step further
10
Multiplying the fraction by $-1$
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(1-x^2\right)+\frac{-1}{x^2+2x+1}\frac{d}{dx}\left(x^2+2x+1\right)$
Intermediate steps
11
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\frac{d}{dx}\left(x^2+2x+1\right)$
Explain this step further
Intermediate steps
12
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)$
Explain this step further
Intermediate steps
13
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(x\right)\right)$
Explain this step further
14
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(-x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+2\right)$
Intermediate steps
15
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=-2\left(\frac{1}{1-x^2}\right)\frac{d}{dx}\left(x^2\right)+\frac{-1}{x^2+2x+1}\left(\frac{d}{dx}\left(x^2\right)+2\right)$
Explain this step further
Intermediate steps
16
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=-2\cdot 2\left(\frac{1}{1-x^2}\right)x+\frac{-1}{x^2+2x+1}\left(2x+2\right)$
Explain this step further
17
Multiply $-2$ times $2$
$\frac{y^{\prime}}{y}=-4\left(\frac{1}{1-x^2}\right)x+\frac{-1}{x^2+2x+1}\left(2x+2\right)$
Intermediate steps
18
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{-4x}{1-x^2}+\frac{-\left(2x+2\right)}{x^2+2x+1}$
Explain this step further
19
Simplify the product $-(2x+2)$
$\frac{y^{\prime}}{y}=\frac{-4x}{1-x^2}+\frac{-2x-2}{x^2+2x+1}$
20
The trinomial $x^2+2x+1$ is a perfect square trinomial, because it's discriminant is equal to zero
$\Delta=b^2-4ac=2^2-4\left(1\right)\left(1\right) = 0$
21
Using the perfect square trinomial formula
$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{x^2}\:and\:b=\sqrt{1}$
22
Factoring the perfect square trinomial
$\frac{y^{\prime}}{y}=\frac{-4x}{1-x^2}+\frac{-2x-2}{\left(x+1\right)^{2}}$
23
Multiply both sides of the equation by $y$
$y^{\prime}=\left(\frac{-4x}{1-x^2}+\frac{-2x-2}{\left(x+1\right)^{2}}\right)y$
24
Substitute $y$ for the original function: $\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
$y^{\prime}=\left(\frac{-4x}{1-x^2}+\frac{-2x-2}{\left(x+1\right)^{2}}\right)\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
25
The derivative of the function results in
$\left(\frac{-4x}{1-x^2}+\frac{-2x-2}{\left(x+1\right)^{2}}\right)\frac{\left(1-x^2\right)^2}{x^2+2x+1}$
Final answer to the exercise $\left(\frac{-4x}{1-x^2}+\frac{-2x-2}{\left(x+1\right)^{2}}\right)\frac{\left(1-x^2\right)^2}{x^2+2x+1}$