If \(f(x)=\sqrt{x^2-4}\) and \(g(x)=3x-2\), then the derivative of \(f(g(x))\) at \(x=3\) is

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Use the chain rule to find the expression for the derivative:

$$ \frac{d}{dx} f(g(x)) = f'(g(x))\cdot g'(x) $$

At \(x=3\),

$$ f'(g(3))\cdot g'(3) $$

Obtaining the equations for \(f'\) and \(g'\):

Using our equations:

$$ g(3)=7 $$ $$ g'(3)=3 $$

Putting it all together:

$$ f'(g(3))\cdot g'(3) $$ $$ = f'(7)\cdot 3 $$ $$ = \frac{7}{\sqrt{45}} \cdot 3 $$ $$ = \frac{7}{3\sqrt{5}}\cdot 3 $$ $$ = \boxed{\frac{7}{\sqrt{5}}} $$