What is the derivative of e^(5ln(tan 5x))e5ln(tan5x)?

1 Answer
May 3, 2018

=25tan^4(5x)sec^2(5x)=25tan4(5x)sec2(5x)

Explanation:

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

Using,

e^(ln(a)eln(a)=a=a

And,

ln(a^x)ln(ax)=x*ln(a)=xln(a)

we get,

e^(5ln(tan(5x))e5ln(tan(5x))
e^(ln(tan(5x))5eln(tan(5x))5
=tan5(5x)=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)