What is the derivative of $f(x)=secx^2tan^2x$ ?

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Answer 1 · 2016-05-06T14:20:56

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

From the given function

$f(x)=sec x^2 tan^2 x$

$f' (x)=sec x^2*d/dx(tan^2 x)+tan^2 x * d/dx(sec x^2)$

$f' (x)=$
$sec x^2*2*tan x*d/dx(tan x)+tan^2 x*sec x^2*tan x^2*d/dx(x^2)$

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

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

God bless....I hope the explanation is useful.

What is the derivative of f(x)=secx^2tan^2x f(x)=secx^2tan^2x?