How do you find the derivative of #f(x)=7^(2x)#?

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Answer 1 · 2018-04-03T23:44:56

#ln49[7^[2x]]#

By the theory of logs #7^[2x# can be written as #e^[2xln7]# ,i.e,

#7^[2x]=e^[2xln7]#.....#[1]#

Therefore, #d/dx7^[2x=## d/[dx]##[e^[2xln7]]#

#d/dx[e^[2xln7]]#=#[e^[2xln7]d/dx[2xln7]]# and since #ln7# is a constant,

#d/dx[2xln7]# = #2ln7#..... So, #d/dx7^[2x#=#2ln7[e^[2xln7]]#.......#[2]#

From .....#[1]#, #e^[2xln7#= #7^[2x# so substituting in 2,

#d/dx 7^[2x#=#2ln7[7^[2x]]#=#ln49[7^[2x]]#. Hope this was helpful.