How do you find the derivative of $y = 3 / (cos2x^2)$ ?

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Answer 1 · 2017-04-14T20:17:19

$dy/dx=12xsec(2x^2)tan(2x^2)$

Rewrite the problem as:

$y=3(sec(2x^2))$

Now take the derivative. You have to use chain rule for $sec(2x^2)$

$dy/dx=3(sec(2x^2))tan(2x^2)(2x^2)'$

$dy/dx=12xsec(2x^2)tan(2x^2)$

How do you find the derivative of y = 3 / (cos2x^2)y = 3 / (cos2x^2)?