What is the derivative of $f(x)=sec^2xcosx$ ?

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Answer 1 · 2016-01-10T14:26:28

$dy/dx = tan(x)sec(x)$

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

But since $sec(x) = 1/cos(x)$ we can simplify that to

$y = sec(x)$ or $y = 1/cos(x)$

from there, using either the product + chain rule or the quotient rule

$dy/dx = d/(du)1/u*d/dxcos(x)$
$dy/dx = -1/u^2 *(-sin(x))$
$dy/dx = sin(x)/cos^2(x)$
$dy/dx = sin(x)/cos(x)*1/cos(x) = tan(x)sec(x)$

What is the derivative of f(x)=sec^2xcosx f(x)=sec^2xcosx?