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If #f(x) =-e^(2x+4) # and #g(x) = tan3x #, what is #f'(g(x)) #?

Answer 1

Charles Arocha

#f'(g(x)) = -2e^(2tan(3x)+4)#

#f'(g(x))# indicates that we need to take the derivative of #f#, and then plug in #g(x)# for #x#.

First, let's find #f'(x)#:

We know that #d/dxe^u = e^u d/dx(u)#

Therefore:

#d/dx(-e^(2x+4)) = -e^(2x+4) * d/dx(2x+4)#

#= -2e^(2x+4)#

So #f'(x) = -2e^(2x+4)#.

We also know that #g(x) = tan(3x)#

Therefore, #f'(g(x)) = -2e^(2tan(3x)+4)#

Final Answer

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Answer 2

Axel Alvarez

To find ( f'(g(x)) ), first, we need to find ( f'(x) ), then substitute ( g(x) ) into ( f'(x) ):

Given ( f(x) = -e^{2x+4} ) and ( g(x) = \tan(3x) ),

( f'(x) = \frac{d}{dx}(-e^{2x+4}) = -2e^{2x+4} ),

Now, we substitute ( g(x) = \tan(3x) ) into ( f'(x) ):

( f'(g(x)) = -2e^{2g(x)+4} = -2e^{2\tan(3x)+4} ).

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