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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

$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
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Step-by-step Solution

Problem to solve:

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

Choose the 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 natural logarithm to both sides of the equality

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

Applying the product rule for logarithms: $\log_b\left(MN\right)=\log_b\left(M\right)+\log_b\left(N\right)$

$\ln\left(y\right)=\ln\left(\left(2x+1\right)^5\right)+\ln\left(\left(x^4-3\right)^6\right)$
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 $\displaystyle f'(x)=\frac{a'}{a}$

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

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

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

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 or more 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 or more functions is the sum of the derivatives of each function

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

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

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

$x+0=x$, 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

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

The derivative of a function multiplied by a constant ($2$) 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 $1$

$10\left(\frac{1}{2x+1}\right)$
16

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

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

Multiplying the fraction by $10$

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

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

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

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

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

$x+0=x$, 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

$y^{\prime}\frac{1}{y}=\frac{10}{2x+1}+6\left(\frac{1}{x^4-3}\right)\frac{d}{dx}\left(x^4\right)$
19

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

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

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

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

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 $4$

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

Multiplying the fraction by $24$

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

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

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

Isolate $y'$

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

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

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

Isolate $y'$

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

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

$y^{\prime}=\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$
23

The derivative of the function results in

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

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, we place the LCM as the denominator of each fraction and in the numerator of each fraction we add the factors that we need to complete

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

Combine and simplify all terms in the same fraction with common denominator $\left(2x+1\right)\left(x^4-3\right)$

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

Multiplying the fraction by $\left(2x+1\right)^5$

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

Multiplying the fraction by $\left(x^4-3\right)^6$

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

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

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

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

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

Multiply the single term $10$ by each term of the polynomial $\left(x^4-3\right)$

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

Simplifying

$\left(10x^4-30+24x^{3}\left(2x+1\right)\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$

Solve the product $24x^{3}\left(2x+1\right)$

$\left(10x^4-30+x^{3}\left(48x+24\right)\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$

Multiply the single term $x^{3}$ by each term of the polynomial $\left(48x+24\right)$

$\left(10x^4-30+48xx^{3}+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$

When multiplying exponents with same base you can add the exponents: $48xx^{3}$

$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
24

Simplifying

$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$

Final Answer

$\left(58x^{4}-30+24x^{3}\right)\left(2x+1\right)^{4}\left(x^4-3\right)^{5}$
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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:

~ 0.38 s

Related topics:

Logarithmic differentiationProperties of LogarithmsBasic Differentiation RulesSum Rule of DifferentiationConstant RulePower Rule for Derivatives

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