The function #f(x) = tan(3^x)# has one zero in the interval #[0, 1.4]#. What is the derivative at this point?

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Answer 1 · 2016-12-01T21:41:54

#pi ln3#

If #tan(3^x) = 0#, then #sin(3^x) = 0# and #cos(3^x) = +-1#

Therefore #3^x# = #kpi# for some integer #k#.

We were told that there is one zero on #[0,1.4]#. That zero is NOT #x=0# (since #tan 1 != 0#). The smallest positive solution must have #3^x = pi#.

Hence, #x = log_3 pi#.

Now let's look at the derivative.

#f'(x) = sec^2(3^x) * 3^x ln3#

We know from above that #3^x = pi#, so at that point

#f' = sec^2(pi) * pi ln3 =(-1)^2 pi ln3 = pi ln3#