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
Apply the formula: $\frac{d}{dx}\left(x\right)$$=y=x$, where $d/dx=\frac{d}{dx}$, $d/dx?x=\frac{d}{dx}\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$ and $x=\left(2x+1\right)^5\left(x^4-3\right)^6$
$y=\left(2x+1\right)^5\left(x^4-3\right)^6$
2
Apply the formula: $y=x$$\to \ln\left(y\right)=\ln\left(x\right)$, where $x=\left(2x+1\right)^5\left(x^4-3\right)^6$
$\ln\left(y\right)=\ln\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$
Intermediate steps
3
Apply the formula: $y=x$$\to y=x$, where $x=\ln\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$ and $y=\ln\left(y\right)$
$\ln\left(y\right)=5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)$
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4
Apply the formula: $\ln\left(y\right)=x$$\to \frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(x\right)$, where $x=5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)$
$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
5
Apply the formula: $\frac{d}{dx}\left(\ln\left(x\right)\right)$$=\frac{1}{x}\frac{d}{dx}\left(x\right)$
$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
6
Apply the formula: $\frac{d}{dx}\left(x\right)$$=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
8
Apply the formula: $\frac{d}{dx}\left(cx\right)$$=c\frac{d}{dx}\left(x\right)$
$\frac{y^{\prime}}{y}=5\frac{d}{dx}\left(\ln\left(2x+1\right)\right)+6\frac{d}{dx}\left(\ln\left(x^4-3\right)\right)$
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Intermediate steps
9
Apply the formula: $\frac{d}{dx}\left(\ln\left(x\right)\right)$$=\frac{1}{x}\frac{d}{dx}\left(x\right)$
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x+1\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
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Intermediate steps
10
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
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Intermediate steps
11
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{y^{\prime}}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
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Intermediate steps
12
Apply the formula: $\frac{d}{dx}\left(nx\right)$$=n\frac{d}{dx}\left(x\right)$, where $n=2$
$\frac{y^{\prime}}{y}=10\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
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13
Apply the formula: $\frac{d}{dx}\left(x\right)$$=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
14
Apply the formula: $a\frac{b}{x}$$=\frac{ab}{x}$
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
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Intermediate steps
15
Apply the formula: $\frac{d}{dx}\left(x^a\right)$$=ax^{\left(a-1\right)}$, where $a=4$
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+6\cdot 4\left(\frac{1}{x^4-3}\right)x^{3}$
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16
Apply the formula: $ab$$=ab$, where $ab=6\cdot 4\left(\frac{1}{x^4-3}\right)x^{3}$, $a=6$ and $b=4$
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+24\left(\frac{1}{x^4-3}\right)x^{3}$
Intermediate steps
17
Apply the formula: $a\frac{b}{x}$$=\frac{ab}{x}$
$\frac{y^{\prime}}{y}=\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}$
Explain this step further
18
Apply the formula: $\frac{a}{b}=c$$\to a=cb$, where $a=y^{\prime}$, $b=y$ and $c=\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}$
$y^{\prime}=\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)y$
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
21
Simplify the derivative
$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
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Final answer to the problem $\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
