What is the derivative of $y = x [sec (3 - 8x)]$ ?

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Answer 1 · 2018-03-18T15:45:11

$1/cos(3-8x)(1-8xtan(3-8x))$

$y = x [sec (3 - 8x)]=x/cos(3-8x)$
$y'=1/cos(3-8x)-8*sin(3-8x)*x/(cos(3-8x)^2)$
$=1/cos(3-8x)(1-8xtan(3-8x))$

What is the derivative of y = x [sec (3 - 8x)] y = x [sec (3 - 8x)]?