Exercise$\frac{d}{dx}\left(2x+1\right)^5\left(x^4-3\right)^6$
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
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
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
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
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
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
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
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
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
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
$\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$
Final answer to the exercise $\left(\frac{10}{2x+1}+\frac{24x^{3}}{x^4-3}\right)\left(2x+1\right)^5\left(x^4-3\right)^6$
