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If #f(x) =cos3x # and #g(x) = sqrt(3x-1 #, what is #f'(g(x)) #?

Answer 1

Charles Arocha

#=-(9/2)sin(3 sqrt(3x-1))/(sqrt(3x-1)#

#f(g(x)#

#=cos (3 g()x))#

#=cos(3 sqrt(3x-1))#

Using chain rule,

#f'=(cos(3 sqrt(3x-1))'#

#=-sin(3 sqrt(3x-1))(3sqrt(3x-1))'#

#=-sin(3 sqrt(3x-1))((-3/2)/(sqrt(3x-1)))(3x-1)'#

#=-sin(3 sqrt(3x-1))((-3/2)/(sqrt(3x-1)))(3)#

#=-(9/2)sin(3 sqrt(3x-1))/(sqrt(3x-1)#

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Answer 2

Luke Alvarez

To find ( f'(g(x)) ), we need to use the chain rule. First, we find the derivative of ( f(x) ) with respect to ( x ), then we substitute ( g(x) ) for ( x ) and find the derivative of ( g(x) ) with respect to ( x ). Finally, we multiply these derivatives together.

First, find ( f'(x) ): [ f'(x) = -3\sin(3x) ]

Then find ( g'(x) ): [ g'(x) = \frac{3}{2\sqrt{3x - 1}} ]

Now substitute ( g(x) ) into ( f'(x) ): [ f'(g(x)) = -3\sin(3g(x)) ]

Multiply ( f'(g(x)) ) by ( g'(x) ): [ f'(g(x)) \cdot g'(x) = -3\sin(3g(x)) \cdot \frac{3}{2\sqrt{3g(x) - 1}} ]

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Answer from HIX Tutor

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