What is the derivative of $y=sec(tan(x))$ ?

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Answer 1 · 2014-07-26T18:10:22

Answer, $y'=sec(tan(x))tan(tan(x))(sec^2(x))$

Solution :

If, $y=sec(f(x))$

then using chain rule, $y'=sec(f(x))tan(f(x))f'(x)$

In same way,

$y'=sec(tan(x))tan(tan(x))(tan(x))'$

$y'=sec(tan(x))tan(tan(x))(sec^2(x))$

What is the derivative of y=sec(tan(x))y=sec(tan(x)) ?