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Step-by-step Solution

Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{6x^2-3}{\sqrt{x^2-1}}\right)$

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Answer

$\frac{12x\sqrt{x^2-1}-\left(x^2-1\right)^{-\frac{1}{2}}x\left(6x^2-3\right)}{x^2-1}$

Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{6x^2-3}{\sqrt{x^2-1}}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

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

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Answer

$\frac{12x\sqrt{x^2-1}-\left(x^2-1\right)^{-\frac{1}{2}}x\left(6x^2-3\right)}{x^2-1}$
$\frac{d}{dx}\left(\frac{6x^2-3}{\sqrt{x^2-1}}\right)$

Main topic:

Quotient rule

Time to solve it:

~ 1.24 seconds

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