How do you find the derivative of $arcsin(4x)$ ?

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Answer 1 · 2016-06-22T14:28:01

$4 / (sqrt{1 - 16x^2})$

$y = arcsin(4x)$

$sin y = 4x$

$cos y y' = 4$

$y' = 4/ (cos y)$

$= 4 / (sqrt{1 - sin^2 y})$

$= 4 / (sqrt{1 - (4x)^2})$

$= 4 / (sqrt{1 - 16x^2})$

How do you find the derivative of arcsin(4x)arcsin(4x)?