# If #f(x) =-sqrt(3x-1) # and #g(x) = (x+3)^3 #, what is #f'(g(x)) #?

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To find ( f'(g(x)) ), we need to find the derivative of ( f ) with respect to ( x ) and then evaluate it at ( g(x) ) and multiply by the derivative of ( g(x) ) with respect to ( x ).

First, find ( f'(x) ): [ f(x) = -\sqrt{3x - 1} ] [ f'(x) = \frac{-1}{2\sqrt{3x - 1}} ]

Now, find ( g'(x) ): [ g(x) = (x + 3)^3 ] [ g'(x) = 3(x + 3)^2 ]

Finally, evaluate ( f'(g(x)) ): [ f'(g(x)) = \frac{-1}{2\sqrt{3g(x) - 1}} ] [ = \frac{-1}{2\sqrt{3((x + 3)^3) - 1}} ] [ = \frac{-1}{2\sqrt{27(x + 3)^2 - 1}} ]

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