How do you find the derivative of Inverse trig function $y= sec^-1 (sqrt(1-x))$ ?

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Answer 1 · 2015-07-02T14:01:10

$y=sec^-1(sqrt(1-x))$
$sec y =sqrt(1-x)$

Now differentiate both sides.

$secy tany dy/dx = 1/sqrt(1-x)*-1$
$dy/dx = -1/(secy tany sqrt(1-x)$

We have $secy = sqrt(1-x)$ calculate

$tany=sqrt(sec^2y-1)$
$tany = sqrt((1-x)-1)$
$tany = sqrt(-x)$

Substitute
$dy/dx = (-1)/(sqrt((1-x)^2(-x)))$

How do you find the derivative of Inverse trig function y= sec^-1 (sqrt(1-x))y= sec^-1 (sqrt(1-x))?