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Find the derivative using logarithmic differentiation method $\frac{d}{dx}\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$

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Solution

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Final Answer

$\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$

Step-by-step Solution

Problem to solve:

$\frac{d}{dx}\left(2x+1\right)^5\left(x^4-3\right)^6$

Solving method

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 logarithm to both sides of the equality

$\ln\left(y\right)=\ln\left(\left(2x+1\right)^5\left(x^4-3\right)^6\right)$
3

Applying the logarithm product rule

$\ln\left(y\right)=\ln\left(\left(2x+1\right)^5\right)+\ln\left(\left(x^4-3\right)^6\right)$
4

Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$

$\ln\left(y\right)=5\ln\left(2x+1\right)+\ln\left(\left(x^4-3\right)^6\right)$
5

Using the power rule of logarithms: $\log_a(x^n)=n\cdot\log_a(x)$

$\ln\left(y\right)=5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)$
6

Derive both sides of the equality with respect to $x$

$\frac{d}{dx}\left(\ln\left(y\right)\right)=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
7

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 $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{y}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$

$1y^{\prime}\left(\frac{1}{y}\right)=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$

Any expression multiplied by $1$ is equal to itself

$y^{\prime}\frac{1}{y}=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
8

The derivative of the linear function is equal to $1$

$y^{\prime}\frac{1}{y}=\frac{d}{dx}\left(5\ln\left(2x+1\right)+6\ln\left(x^4-3\right)\right)$
9

The derivative of a sum of two functions is the sum of the derivatives of each function

$y^{\prime}\frac{1}{y}=\frac{d}{dx}\left(5\ln\left(2x+1\right)\right)+\frac{d}{dx}\left(6\ln\left(x^4-3\right)\right)$
10

The derivative of a function multiplied by a constant ($5$) is equal to the constant times the derivative of the function

$y^{\prime}\frac{1}{y}=5\frac{d}{dx}\left(\ln\left(2x+1\right)\right)+\frac{d}{dx}\left(6\ln\left(x^4-3\right)\right)$
11

The derivative of a function multiplied by a constant ($6$) is equal to the constant times the derivative of the function

$y^{\prime}\frac{1}{y}=5\frac{d}{dx}\left(\ln\left(2x+1\right)\right)+6\frac{d}{dx}\left(\ln\left(x^4-3\right)\right)$
12

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 $\displaystyle f'(x)=\frac{a'}{a}$

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x+1\right)+6\frac{d}{dx}\left(\ln\left(x^4-3\right)\right)$
13

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 $\displaystyle f'(x)=\frac{a'}{a}$

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x+1\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
14

The derivative of a sum of two functions is the sum of the derivatives of each function

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\left(\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right)\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\left(\frac{d}{dx}\left(2x\right)+0\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

$x+0=x$, where $x$ is any expression

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
15

The derivative of the constant function ($1$) is equal to zero

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

$y^{\prime}\frac{1}{y}=5\cdot 2\left(\frac{1}{2x+1}\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

Multiply $5$ times $2$

$y^{\prime}\frac{1}{y}=10\left(\frac{1}{2x+1}\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

Multiply the fraction and term

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

$\frac{10}{2x+1}\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$\frac{10}{2x+1}$
16

The derivative of the linear function times a constant, is equal to the constant

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$
17

The derivative of a sum of two functions is the sum of the derivatives of each function

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\left(\frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(-3\right)\right)$

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\left(\frac{d}{dx}\left(2x\right)+0\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

$x+0=x$, where $x$ is any expression

$y^{\prime}\frac{1}{y}=5\left(\frac{1}{2x+1}\right)\frac{d}{dx}\left(2x\right)+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4-3\right)$

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\left(\frac{d}{dx}\left(x^4\right)+0\right)$

$x+0=x$, where $x$ is any expression

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
18

The derivative of the constant function ($-3$) is equal to zero

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\cdot 4x^{\left(4-1\right)}\left(\frac{1}{x^4-3}\right)$

Subtract the values $4$ and $-1$

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\cdot 4x^{3}\left(\frac{1}{x^4-3}\right)$

Multiply $6$ times $4$

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+24x^{3}\left(\frac{1}{x^4-3}\right)$

Multiply the fraction and term

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}$
19

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}$

$y^{\prime}=\frac{\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}}{\frac{1}{y}}$

Divide fractions $\frac{\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}}{\frac{1}{y}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$y^{\prime}=y\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)$
20

Isolate $y'$

$y^{\prime}=y\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)$
21

Substitute $y$ for the original function: $\left(2x+1\right)^5\left(x^4-3\right)^6$

$y^{\prime}=\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$
22

The derivative of the function results in

$\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$

Final Answer

$\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$

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$\frac{d}{dx}\left(2x+1\right)^5\left(x^4-3\right)^6$

Main topic:

Logarithmic differentiation

Related Formulas:

6. See formulas

Time to solve it:

~ 0.12 s

Related topics:

Logarithmic differentiationProperties of LogarithmsBasic Differentiation RulesSum Rule of DifferentiationConstant RulePower rule for derivatives

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