What is the equation of the tangent line of #f(x)=ln(2x)-3e^(x^2)# at #x=4e#?

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Answer 1 · 2017-11-29T00:42:09

Using point-slope form, the tangent line is:

#y-(ln(8e)-3e^(16e^2)) = (1/(4e)-24e^(16e^2+1))(x-4e)#

First find #f(4e) = ln(2*4e)-3e^((4e)^2) = ln(8e)-3e^(16e^2)#.

Now find #f'(x)# using the Chain Rule a few places:

#f'(x) = 2/(2x)-3e^(x^2)*2x=1/x-6xe^(x^2)#.

Now find #f'(4e)#.

#f'(4e) = 1/(4e)-6(4e)e^((4e)^2)=1/(4e)-24e^(16e^2+1)#.

Using point-slope form, the tangent line is:

#y-(ln(8e)-3e^(16e^2)) = (1/(4e)-24e^(16e^2+1))(x-4e)#