What is the derivative of #f(x)=sec^-1(x)# ?
1 Answer
Process:
First, we will make the equation a little easier to deal with. Take the secant of both sides:
#y = sec^-1 x#
#sec y = x#
Next, rewrite in terms of
#1/cos y = x#
And solve for
#1 = xcosy#
#1/x = cosy#
#y = arccos(1/x)#
Now this looks much easier to differentiate. We know that
so we can use this identity as well as the chain rule:
#dy/dx = -1/sqrt(1 - (1/x)^2) * d/dx[1/x]#
A bit of simplification:
#dy/dx = -1/sqrt(1 - 1/x^2) * (-1/x^2)#
A little more simplification:
#dy/dx = 1/(x^2sqrt(1 - 1/x^2))#
To make the equation a little prettier I will move the
#dy/dx = 1/(sqrt(x^4(1 - 1/x^2)))#
Some final reduction:
#dy/dx = 1/(sqrt(x^4 - x^2))#
And there's our derivative.
When differentiating inverse trig functions, the key is getting them in a form that's easy to deal with. More than anything, they're an exercise in your knowledge of trig identities and algebraic manipulation.
