Question #3063d

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Answer 1 · 2016-04-23T21:35:54

#5#

The chain rule states that

#d/dxf(g(x))=f'(g(x))*g'(x)#

Thus,

#d/dxf(e^tanx)=f'(e^tanx)*d/dx(e^tanx)#

#=f'(e^tanx) * e^tanx * sec^2x#

Note that the chain rule was also used to find #d/dx(e^tanx)#.

So, if we want to find the derivative when #x=0#, plug in #0# for #x#:

#d/dxf(e^tanx)|_(x=0)=f'(e^tan0) * e^tan0 * sec^2 0#

#=f'(e^0) * e^0 * (1)^2#

#=f'(1) * 1 * 1#

#=5#

Recall that #f'(1)=5# was given in the question.