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If #f(x) =-e^(-3x-7) # and #g(x) = -2sec^2x #, what is #f'(g(x)) #?

Answer 1

Isaiah Allen

#=12e^(-7)e^(6sec^2x)sec^2xtanx#.

#f(g)=-e^(-3g-7)=e^(-7)e^(6sec^2x)#

So, #f'(g)=(df)/(dx)#

#=e^(-7)e(6sec^2x)(6sec^2x)'#

#=e^(-7)e(6sec^2x)12secx(secx)'#

#=12e^(-7)e^(6sec^2x)sec^2xtanx#.

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Answer 2

Emily Ammerman

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

Given ( f(x) = -e^{-3x-7} ), we find ( f'(x) = 3e^{-3x-7} ).

Now, substituting ( g(x) ) into ( f'(x) ), we have:

[ f'(g(x)) = 3e^{-3(g(x))-7} ]

Substitute ( g(x) = -2\sec^2(x) ) into the expression:

[ f'(g(x)) = 3e^{-3(-2\sec^2(x))-7} ]

Therefore, ( f'(g(x)) = 3e^{6\sec^2(x)-7} ).

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