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

Answer 1

Emma Aldridge

#f(x)=sin^3x# , #D_f=RR#

#g(x)=sqrt(3x-1)#, #Dg=[1/3,+oo)#

#D_(fog)={##AAx##in##RR:##x##in##D_g#, #g(x)##in##D_f}#

#x>=1/3# , #sqrt(3x-1)##in##RR# #-># #x##in##[1/3,+oo)#

#AAx##in##[1/3,+oo)#,

#f'(x)=3sin^2x(sinx)'=3sin^2xcosx#

so #(fog)'(x)=sin^2(sqrt(3x-1))cos(sqrt(3x-1))*9/(2sqrt(3x-1))#

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Answer 2

Isabella Ackerman

To find ( f'(g(x)) ), we need to use the chain rule.

First, we find ( f'(x) ): [ f(x) = \sin^3(x) ] [ f'(x) = 3\sin^2(x) \cdot \cos(x) ]

Now, we substitute ( g(x) ) into ( f'(x) ): [ f'(g(x)) = 3\sin^2(g(x)) \cdot \cos(g(x)) ]

Since ( g(x) = \sqrt{3x-1} ), we have: [ \sin(g(x)) = \sin(\sqrt{3x-1}) ]

Thus, [ f'(g(x)) = 3\sin^2(\sqrt{3x-1}) \cdot \cos(\sqrt{3x-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.