Final answer to the problem
$\left(\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-2x-6}{\left(x+2\right)\left(x+4\right)}+\frac{-1}{x-1}\right)\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}$
Got another answer? Verify it here!
Step-by-step Solution
Specify the solving method
Find the derivative Find the derivative using the product rule Find the derivative using the quotient rule Find the derivative using logarithmic differentiation Find the derivative using the definition Suggest another method or feature
Send
Intermediate steps
1
Simplify the derivative by applying the properties of logarithms
$\frac{d}{dx}\left(\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}\right)$
Explain this step further
2
To derive the function $\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}$
3
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}\right)$
Intermediate steps
4
Apply logarithm properties to both sides of the equality
$\ln\left(y\right)=\ln\left(x^2-16\right)+\ln\left(x^2+2x-3\right)-\ln\left(x^2+6x+8\right)-\ln\left(x-1\right)$
Explain this step further
5
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(\ln\left(x^2-16\right)+\ln\left(x^2+2x-3\right)-\ln\left(x^2+6x+8\right)-\ln\left(x-1\right)\right)$
6
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln\left(x^2-16\right)+\ln\left(x^2+2x-3\right)-\ln\left(x^2+6x+8\right)-\ln\left(x-1\right)\right)$
Intermediate steps
7
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(x^2-16\right)+\ln\left(x^2+2x-3\right)-\ln\left(x^2+6x+8\right)-\ln\left(x-1\right)\right)$
Explain this step further
8
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(x^2-16\right)\right)+\frac{d}{dx}\left(\ln\left(x^2+2x-3\right)\right)+\frac{d}{dx}\left(-\ln\left(x^2+6x+8\right)\right)+\frac{d}{dx}\left(-\ln\left(x-1\right)\right)$
Intermediate steps
9
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(\ln\left(x^2-16\right)\right)+\frac{d}{dx}\left(\ln\left(x^2+2x-3\right)\right)-\frac{d}{dx}\left(\ln\left(x^2+6x+8\right)\right)-\frac{d}{dx}\left(\ln\left(x-1\right)\right)$
Explain this step further
Intermediate steps
10
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2-16\right)+\frac{1}{x^2+2x-3}\frac{d}{dx}\left(x^2+2x-3\right)-\left(\frac{1}{x^2+6x+8}\right)\frac{d}{dx}\left(x^2+6x+8\right)-\left(\frac{1}{x-1}\right)\frac{d}{dx}\left(x-1\right)$
Explain this step further
11
Multiplying the fraction by $-1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2-16\right)+\frac{1}{x^2+2x-3}\frac{d}{dx}\left(x^2+2x-3\right)+\frac{-1}{x^2+6x+8}\frac{d}{dx}\left(x^2+6x+8\right)-\left(\frac{1}{x-1}\right)\frac{d}{dx}\left(x-1\right)$
12
Multiplying the fraction by $-1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2-16\right)+\frac{1}{x^2+2x-3}\frac{d}{dx}\left(x^2+2x-3\right)+\frac{-1}{x^2+6x+8}\frac{d}{dx}\left(x^2+6x+8\right)+\frac{-1}{x-1}\frac{d}{dx}\left(x-1\right)$
13
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(-16\right)\right)+\frac{1}{x^2+2x-3}\frac{d}{dx}\left(x^2+2x-3\right)+\frac{-1}{x^2+6x+8}\frac{d}{dx}\left(x^2+6x+8\right)+\frac{-1}{x-1}\frac{d}{dx}\left(x-1\right)$
14
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(-16\right)\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(-3\right)\right)+\frac{-1}{x^2+6x+8}\frac{d}{dx}\left(x^2+6x+8\right)+\frac{-1}{x-1}\frac{d}{dx}\left(x-1\right)$
15
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(-16\right)\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(-3\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)+\frac{d}{dx}\left(8\right)\right)+\frac{-1}{x-1}\frac{d}{dx}\left(x-1\right)$
16
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(-16\right)\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(-3\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)+\frac{d}{dx}\left(8\right)\right)+\frac{-1}{x-1}\left(\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-1\right)\right)$
17
The derivative of the constant function ($-16$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(-3\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)+\frac{d}{dx}\left(8\right)\right)+\frac{-1}{x-1}\left(\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-1\right)\right)$
18
The derivative of the constant function ($-3$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)+\frac{d}{dx}\left(8\right)\right)+\frac{-1}{x-1}\left(\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-1\right)\right)$
19
The derivative of the constant function ($8$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)\right)+\frac{-1}{x-1}\left(\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-1\right)\right)$
20
The derivative of the constant function ($-1$) is equal to zero
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)\right)+\frac{-1}{x-1}\frac{d}{dx}\left(x\right)$
Intermediate steps
21
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)\right)+\frac{-1}{x-1}$
Explain this step further
Intermediate steps
22
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(x\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(6x\right)\right)+\frac{-1}{x-1}$
Explain this step further
Intermediate steps
23
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(x\right)\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+6\frac{d}{dx}\left(x\right)\right)+\frac{-1}{x-1}$
Explain this step further
Intermediate steps
24
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+2\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+6\frac{d}{dx}\left(x\right)\right)+\frac{-1}{x-1}$
Explain this step further
Intermediate steps
25
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{1}{x^2-16}\frac{d}{dx}\left(x^2\right)+\frac{1}{x^2+2x-3}\left(\frac{d}{dx}\left(x^2\right)+2\right)+\frac{-1}{x^2+6x+8}\left(\frac{d}{dx}\left(x^2\right)+6\right)+\frac{-1}{x-1}$
Explain this step further
Intermediate steps
26
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=2\left(\frac{1}{x^2-16}\right)x+\frac{1}{x^2+2x-3}\left(2x+2\right)+\frac{-1}{x^2+6x+8}\left(2x+6\right)+\frac{-1}{x-1}$
Explain this step further
Intermediate steps
27
Multiply the fraction and term
$\frac{y^{\prime}}{y}=\frac{2x}{x^2-16}+\frac{2x+2}{x^2+2x-3}+\frac{-1}{x^2+6x+8}\left(2x+6\right)+\frac{-1}{x-1}$
Explain this step further
28
Multiplying the fraction by $2x+6$
$\frac{y^{\prime}}{y}=\frac{2x}{x^2-16}+\frac{2x+2}{x^2+2x-3}+\frac{-\left(2x+6\right)}{x^2+6x+8}+\frac{-1}{x-1}$
29
Factor the trinomial $x^2+2x-3$ finding two numbers that multiply to form $-3$ and added form $2$
$\begin{matrix}\left(-1\right)\left(3\right)=-3\\ \left(-1\right)+\left(3\right)=2\end{matrix}$
$\frac{y^{\prime}}{y}=\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-\left(2x+6\right)}{x^2+6x+8}+\frac{-1}{x-1}$
31
Factor the trinomial $x^2+6x+8$ finding two numbers that multiply to form $8$ and added form $6$
$\begin{matrix}\left(2\right)\left(4\right)=8\\ \left(2\right)+\left(4\right)=6\end{matrix}$
$\frac{y^{\prime}}{y}=\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-\left(2x+6\right)}{\left(x+2\right)\left(x+4\right)}+\frac{-1}{x-1}$
33
Simplify the product $-(2x+6)$
$\frac{y^{\prime}}{y}=\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-2x-6}{\left(x+2\right)\left(x+4\right)}+\frac{-1}{x-1}$
34
Multiply both sides of the equation by $y$
$y^{\prime}=\left(\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-2x-6}{\left(x+2\right)\left(x+4\right)}+\frac{-1}{x-1}\right)y$
35
Substitute $y$ for the original function: $\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}$
$y^{\prime}=\left(\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-2x-6}{\left(x+2\right)\left(x+4\right)}+\frac{-1}{x-1}\right)\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}$
36
The derivative of the function results in
$\left(\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-2x-6}{\left(x+2\right)\left(x+4\right)}+\frac{-1}{x-1}\right)\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}$
Final answer to the problem$\left(\frac{2x}{x^2-16}+\frac{2x+2}{\left(x-1\right)\left(x+3\right)}+\frac{-2x-6}{\left(x+2\right)\left(x+4\right)}+\frac{-1}{x-1}\right)\frac{\left(x^2-16\right)\left(x^2+2x-3\right)}{\left(x^2+6x+8\right)\left(x-1\right)}$