What's the derivative of $arctan(6^x)$ ?

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Answer 1 · 2016-06-24T18:21:31

$y' = ( ln(6) 6^x)/( 1 + 6^{2x} )$

$y=arctan(6^x)$
$tan y=6^x$
$sec^2 y \ y' = (6^x)'$

$z = 6^x$
$ln z = x ln(6)$
$1/z z' = ln(6)$
$z' = ln(6) 6^x$

$\implies sec^2 y \ y' = ln(6) 6^x$

$y' = ( ln(6) 6^x)/( sec^2 y)$

using $tan^2 +1 = sec^2$

$y' = ( ln(6) 6^x)/( 1 + 6^{2x} )$

What's the derivative of arctan(6^x)arctan(6^x)?