Final answer to the problem
$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
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Step-by-step Solution
1
To derive the function $\left(2x+1\right)^5\left(x^4-3\right)^6$, use the method of logarithmic differentiation. First, assign the function to $y$, then take the natural logarithm of both sides of the equation
$y=\left(2x+1\right)^5\left(x^4-3\right)^6$
2
Apply natural logarithm to both sides of the equality
$\ln\left(y\right)=\ln\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$
Intermediate steps
Apply logarithm properties to both sides of the equality
$\ln\left(y\right)=\ln\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$
Applying the product rule for logarithms: $\log_b\left(MN\right)=\log_b\left(M\right)+\log_b\left(N\right)$
$\ln\left(y\right)=\ln\left(\left(2x+1\right)^5\right)+\ln\left(\left(x^4-3\right)^6\right)$
Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$
$\ln\left(y\right)=5\ln\left(2x+1\right)+\ln\left(\left(x^4-3\right)^6\right)$
Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$
$\ln\left(y\right)=5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)$
3
Apply logarithm properties to both sides of the equality
$\ln\left(y\right)=5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)$
Explain this step further
4
Derive both sides of the equality with respect to $x$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
6
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
7
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=\frac{d}{dx}\left(5\ln\left(2x+1\right)\right)+\frac{d}{dx}\left(6\ln\left(x^4-3\right)\right)$
Intermediate steps
The derivative of a function multiplied by a constant ($6$) is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=5\frac{d}{dx}\left(\ln\left(2x+1\right)\right)+6\frac{d}{dx}\left(\ln\left(x^4-3\right)\right)$
8
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{y^{\prime}}{y}=5\frac{d}{dx}\left(\ln\left(2x+1\right)\right)+6\frac{d}{dx}\left(\ln\left(x^4-3\right)\right)$
Explain this step further
Intermediate steps
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x+1\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x+1\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
Explain this step further
Intermediate steps
The derivative of the constant function ($1$) is equal to zero
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
10
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
Explain this step further
Intermediate steps
The derivative of the constant function ($-3$) is equal to zero
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
11
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
Explain this step further
Intermediate steps
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$10\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(x\right)$
The derivative of the linear function is equal to $1$
$10\left(\frac{1}{2x+1}\right)$
12
The derivative of the linear function times a constant, is equal to the constant
$\frac{y^{\prime}}{y}=10\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
Explain this step further
13
The derivative of the linear function is equal to $1$
$\frac{y^{\prime}}{y}=10\left(\frac{1}{2x+1}\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
Intermediate steps
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{10\cdot 1}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
14
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
Explain this step further
Intermediate steps
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$24\left(\frac{1}{x^4-3}\right)x^{\left(4-1\right)}$
Subtract the values $4$ and $-1$
$24\left(\frac{1}{x^4-3}\right)x^{3}$
15
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+6\cdot 4\left(\frac{1}{x^4-3}\right)x^{3}$
Explain this step further
16
Multiply $6$ times $4$
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+24\left(\frac{1}{x^4-3}\right)x^{3}$
Intermediate steps
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+\frac{24\cdot 1x^{3}}{x^4-3}$
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{10\cdot 1}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
Any expression multiplied by $1$ is equal to itself
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}$
17
Multiply the fraction by the term
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}$
Explain this step further
18
Multiply both sides of the equation by $y$
$y^{\prime}=\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)y$
19
Substitute $y$ for the original function: $\left(2x+1\right)^5\left(x^4-3\right)^6$
$y^{\prime}=\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$
20
The derivative of the function results in
$\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$
Intermediate steps
The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors
$L.C.M.=\left(2x+1\right)\left(x^4-3\right)$
Obtained the least common multiple (LCM), we place it as the denominator of each fraction, and in the numerator of each fraction we add the factors that we need to complete
$\frac{10\left(x^4-3\right)}{\left(2x+1\right)\left(x^4-3\right)}+\frac{24x^{3}\left(2x+1\right)}{\left(2x+1\right)\left(x^4-3\right)}$
$\frac{10x^4-30}{\left(2x+1\right)\left(x^4-3\right)}+\frac{48x^{3}x+24x^{3}}{\left(2x+1\right)\left(x^4-3\right)}$
Combine and simplify all terms in the same fraction with common denominator $\left(2x+1\right)\left(x^4-3\right)$
$\frac{58x^{4}-30+24x^{3}}{\left(2x+1\right)\left(x^4-3\right)}\left(2x+1\right)^5\left(x^4-3\right)^6$
Multiplying the fraction by $\left(2x+1\right)^5\left(x^4-3\right)^6$
$\frac{\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6}{\left(2x+1\right)\left(x^4-3\right)}$
Simplify the fraction $\frac{\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6}{\left(2x+1\right)\left(x^4-3\right)}$ by $2x+1$
$\frac{\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^6}{x^4-3}$
Simplify the fraction $\frac{\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^6}{x^4-3}$ by $x^4-3$
$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
21
Simplify the derivative
$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
Explain this step further
Final answer to the problem $\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
