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