What is the derivative of #e^(5ln(tan 5x))#?

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Answer 1 · 2018-05-03T10:50:16

#=25tan^4(5x)sec^2(5x)#

EDIT: Sorry, I didn't catch that you wanted the derivative. Had to come back to redo it.

Using,

#e^(ln(a)##=a#

And,

#ln(a^x)##=x*ln(a)#

we get,

#e^(5ln(tan(5x))#
#e^(ln(tan(5x))5#
#=tan5(5x)#

from there, we can use the chain rule

#(u^5)'*(tan(5x))'#

where

#(tan(5x)) = sec^2(5x)*5#

which gives,

#5u^4sec^2(5x)*5#

In total that becomes,

#25tan^4(5x)sec^2(5x)#