#F( x ) = e ^ { 3\ln ( x ^ { 2} - 1) } #, what is # f ^ { \prime } ( 1)#?

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Answer 1 · 2017-10-09T00:44:58

#f'(1) = 0#

Begin by rewriting the exponent. You can take advantage of a rule for logarithms:

#b ln a = ln a^b#

Thus:

#F(x) = e^(3 ln (x^2 - 1)) = e^(ln (x^2-1)^3)#

Next, we can take advantage of the fact that the exponential and natural logarithm are inverses:

#e^(ln (x^2-1)^3) = (x^2-1)^3#

Now we can proceed with finding the derivative, and then evaluate it at 1:

#f'(x) = 3(x^2-1)^2 * 2x = 6x(x^2-1)^2#

#f'(1) = 6(1)(1^2-1)^2 = 6(0) = 0#