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If #f(x) =cos3x # and #g(x) = (2x-1)^2 #, what is #f'(g(x)) #?

Answer 1

Paisley Johnson

#(df)/(dg)=(-3sin3x)/(4(2x-1))#

As per chain formula #(df)/(dx)=(df)/(dg)xx(dg)/(dx)#. hence

#(df)/(dg)=((df)/(dx))/((dg)/(dx))#

As #f(x)=sin3x# and #g(x)=(2x-1)^2#

#(df)/(dx)=-sin3x xx3# and #(dg)/(dx)=2(2x-1)xx2#

Hence #(df)/(dg)=(-3sin3x)/(4(2x-1))#

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Answer 2

Emily Ammerman

To find ( f'(g(x)) ), we first need to find the derivative of ( f(x) ) and then substitute ( g(x) ) into it.

Given ( f(x) = \cos(3x) ) and ( g(x) = (2x - 1)^2 ):

Find ( f'(x) ): ( f'(x) = -3\sin(3x) )
Substitute ( g(x) ) into ( f'(x) ): ( f'(g(x)) = -3\sin(3(2x - 1)^2) )

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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.