Final Answer
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$
Step-by-step explanation
$\frac{d}{dx}\left(2x+1\right)^5\left(x^4-3\right)^6$
Choose the solving method
1
Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=\left(2x+1\right)^5$ and $g=\left(x^4-3\right)^6$
$\frac{d}{dx}\left(\left(2x+1\right)^5\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
2
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{\left(5-1\right)}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
3
Subtract the values $5$ and $-1$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+\left(2x+1\right)^5\frac{d}{dx}\left(\left(x^4-3\right)^6\right)$
Intermediate steps
4
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x+1\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
5
The derivative of a sum of two functions is the sum of the derivatives of each function
$5\left(2x+1\right)^{4}\left(\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right)\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
Intermediate steps
6
The derivative of the constant function ($1$) is equal to zero
$5\left(2x+1\right)^{4}\frac{d}{dx}\left(2x\right)\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
Intermediate steps
7
The derivative of the linear function times a constant, is equal to the constant
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
8
The derivative of a sum of two functions is the sum of the derivatives of each function
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\left(\frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(-3\right)\right)$
Intermediate steps
9
The derivative of the constant function ($-3$) is equal to zero
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
Intermediate steps
10
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$
Final Answer
$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$