What is the derivative of f(x)=sec^-1(x)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 xy=sec−1x
sec y = xsecy=x
Next, rewrite in terms of
1/cos y = x1cosy=x
And solve for
1 = xcosy1=xcosy
1/x = cosy1x=cosy
y = arccos(1/x)y=arccos(1x)
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]dydx=−1√1−(1x)2⋅ddx[1x]
A bit of simplification:
dy/dx = -1/sqrt(1 - 1/x^2) * (-1/x^2)dydx=−1√1−1x2⋅(−1x2)
A little more simplification:
dy/dx = 1/(x^2sqrt(1 - 1/x^2))dydx=1x2√1−1x2
To make the equation a little prettier I will move the
dy/dx = 1/(sqrt(x^4(1 - 1/x^2)))dydx=1√x4(1−1x2)
Some final reduction:
dy/dx = 1/(sqrt(x^4 - x^2))dydx=1√x4−x2
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.
