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Find the derivative of $\left(2x+1\right)^5\left(x^4-3\right)^6$

Step-by-step Solution

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Solution

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Final Answer

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

Problem to solve:

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

Specify 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 $g=\left(x^4-3\right)^6$

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

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

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

Subtract the values $5$ and $-1$

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

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

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

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

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

Subtract the values $5$ and $-1$

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

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

Subtract the values $6$ and $-1$

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

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

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

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

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

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

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

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

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

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

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

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

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4-3\right)$
7

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

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

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

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

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

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

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\left(\frac{d}{dx}\left(x^4\right)+0\right)$

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

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+6\left(2x+1\right)^5\left(x^4-3\right)^{5}\frac{d}{dx}\left(x^4\right)$
8

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

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

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

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

Subtract the values $4$ and $-1$

$10\left(2x+1\right)^{4}\left(x^4-3\right)^6+24x^{3}\left(2x+1\right)^5\left(x^4-3\right)^{5}$
9

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

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

asin
acos
atan
acot
asec
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sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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

Main topic:

Differential Calculus

Used formulas:

6. See formulas

Time to solve it:

~ 0.07 s

Related topics:

Differential CalculusProduct Rule of differentiationBasic Differentiation RulesPower Rule for DerivativesSum Rule of DifferentiationConstant Rule for Differentiation

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