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

1 Answer
Manikandan S.
Jul 2, 2015

Explanation:

#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)))#

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