Final answer to the problem
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
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Step-by-step Solution
1
Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\left(2x+1\right)^5$ and $g=\left(x^4-3\right)^6$
$\frac{d}{dx}\left(\left(2x+1\right)^5\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Intermediate steps
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Add the values $5$ and $-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
Subtract the values $5$ and $-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
2
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
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 $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{6-1}\frac{d}{dx}\left(x^4-3\right)$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Add the values $5$ and $-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{5-1}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
Subtract the values $5$ and $-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6$
Add the values $6$ and $-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$6\left(2x+1\right)^5\left(x^4-3\right)^{6-1}\frac{d}{dx}\left(x^4-3\right)$
Subtract the values $6$ and $-1$
$6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
3
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
Explain this step further
Intermediate steps
The derivative of the constant function ($1$) is equal to zero
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
4
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
Explain this step further
Intermediate steps
The derivative of the constant function ($-3$) is equal to zero
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
5
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
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
$10\left(2x+1\right)^{4}\frac{d}{dx}\left(x\right)\left(x^4-3\right)^6$
The derivative of the linear function is equal to $1$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6$
6
The derivative of the linear function times a constant, is equal to the constant
$10\left(2x+1\right)^{4}\frac{d}{dx}\left(x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
Explain this step further
7
The derivative of the linear function is equal to $1$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
Intermediate steps
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{\left(4-1\right)}$
Subtract the values $4$ and $-1$
$24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
8
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\cdot 4\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
Explain this step further
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
Final answer to the problem $10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24\left(2x+1\right)^5\left(x^4-3\right)^{5}x^{3}$
