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

Answer 1

Aurora Alvarado

#f' (g(x)) =( -9(x+3)^2)/(2 sqrt(3(x+3)^3 -1)#

#f(x) = -sqrt(3x-1)# and #g(x)=(x+3)^3#

#f(g(x)) = -sqrt(3(g(x))-1)#

#f(g(x))= -sqrt(3(x+3)^3 -1)#

#f' (g(x)) = (-1)*(1/(2 sqrt(3(x+3)^3 -1)))*3*3(x+3)^2*1#

#f' (g(x)) =( -9(x+3)^2)/(2 sqrt(3(x+3)^3 -1)#

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Answer 2

Isaiah Allen

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.