What is the derivative of $y = {sec}^{2} x + {tan}^{2} x$ $y=sec^2 x + tan^2 x$ ?

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Answer 1 · 2018-06-23T12:20:09

$f'(x)=2sec^2(x)tan(x)+2tan(x)*(1+tan^2(x))$

Note that

$(sec(x))'=sec(x)tan(x)$
$(tan(x))'=1+tan^2(x)$

Answer 2 · 2018-06-23T17:08:03

Shown below...

Another way of thinking about it:

$1 + tan^2 x = sec^2 x$

$=>1 + tan^2x + tan^2 x = sec^2 x + tan^2 x$

$=> 1 + 2tan^2 x = sec^2 x + tan^2 x$

$=> d/dx ( 1 + 2 tan^2 x )$

Use $d/dx tanx = sec^2 x$

$= 2*2tanx * d/dx tanx$

$= 4tanx sec^2x$

What is the derivative of y=sec^2 x + tan^2 xy=sec2x+tan2xy=sec^2 x + tan^2 x?