Answer
$y'=5\sec5x$
Work Step by Step
$y=\ln|\sec5x+\tan5x|$
Start the differentiation process by using the chain rule:
$y'=\dfrac{(|\sec5x+\tan5x|)'}{|\sec5x+\tan5x|}=...$
Apply the chain rule again to find the derivative indicated in the numerator:
$...=\dfrac{\Big(\dfrac{\sec5x+\tan5x}{|\sec5x+\tan5x|}\Big)(\sec5x+\tan5x)'}{|\sec5x+\tan5x|}=...$
$...=\dfrac{(\sec5x+\tan5x)(\sec5x+\tan5x)'}{|\sec5x+\tan5x|^{2}}=...$
Use the chain rule one last time to evaluate the derivative indicated in the numerator and simplify:
$...=\dfrac{(\sec5x+\tan5x)[(\sec5x\tan5x)(5x)'+(5x)'\sec^{2}5x]}{|\sec5x+\tan5x|^{2}}=...$
$...=\dfrac{(\sec5x+\tan5x)[5(\sec5x\tan5x)+5\sec^{2}5x]}{|\sec5x+\tan5x|^{2}}=...$
$...=\dfrac{5(\sec5x+\tan5x)(\sec5x\tan5x+\sec^{2}5x)}{|\sec5x+\tan5x|^{2}}=...$
Take out common factor $\sec5x$ from the numerator:
$...=\dfrac{5\sec5x(\sec5x+\tan5x)(\sec5x+\tan5x)}{|\sec5x+\tan5x|^{2}}=...$
$...=\dfrac{5\sec5x(\sec5x+\tan5x)^{2}}{|\sec5x+\tan5x|^{2}}$
=$5\sec5x$
