# Step-by-step Solution

## Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{-8x^{-5}+6}{-4x^{-6}+8}\right)$

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$\frac{40x^{-6}\left(-4x^{-6}+8\right)-24x^{-7}\left(-8x^{-5}+6\right)}{\left(-4x^{-6}+8\right)^2}$

## Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{-8x^{-5}+6}{-4x^{-6}+8}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

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

$\frac{40x^{-6}\left(-4x^{-6}+8\right)-24x^{-7}\left(-8x^{-5}+6\right)}{\left(-4x^{-6}+8\right)^2}$
$\frac{d}{dx}\left(\frac{-8x^{-5}+6}{-4x^{-6}+8}\right)$

Quotient rule

~ 0.68 seconds

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