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Step-by-step Solution

Find the derivative of $\frac{1}{x^2-3x-1}$ using the constant rule

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Answer

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

Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{1}{x^2-3x-1}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

The derivative of the constant function is equal to zero

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

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Answer

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

Main topic:

Constant rule

Used formulas:

6. See formulas

Time to solve it:

~ 0.61 seconds