What is the derivative of $e^{5 ln (tan 5 x)}$ $e^(5ln(tan 5x))$ ?

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Answer 1 · 2018-05-03T10:50:16

$= 25 {tan}^{4} (5 x) {sec}^{2} (5 x)$ $=25tan^4(5x)sec^2(5x)$

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

Using,

$e^{ln (a)}$ $e^(ln(a)$ $= a$ $=a$

And,

$ln (a^{x})$ $ln(a^x)$ $= x \cdot ln (a)$ $=x*ln(a)$

we get,

$e^{5 ln (tan (5 x))}$ $e^(5ln(tan(5x))$
$e^{ln (tan (5 x)) 5}$ $e^(ln(tan(5x))5$
$= tan 5 (5 x)$ $=tan5(5x)$

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,

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

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