How do you find the derivative of $y=ln(cosx^2)$ ?

Question

11613 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-01T14:56:49

$y = ln(cosx^2)$
$e^y=cos(x^2)$
$y'e^y=-2xsin(x^2)$
$y' = (-2xsin(x^2))/e^y$

but $e^y =cos(x^2)$

so

$y' = (-2xsin(x^2))/cos(x^2)$

$y' = -2xtan(x^2)$

How do you find the derivative of y=ln(cosx^2)y=ln(cosx^2)?