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Step-by-step Solution

Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{\left(x^2+2x+6\right)^4}{x+1}\right)$

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Answer

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

Step-by-step explanation

Problem to solve:

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

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

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

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Answer

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

Main topic:

Quotient rule of differentiation

Used formulas:

6. See formulas

Time to solve it:

~ 0.89 seconds