Derivatives of inverse functions
%20%3D%20%5Carctan(x)#card=math&code=y%20%3D%20%5Ctan%5E%7B-1%7D%28x%29%20%3D%20%5Carctan%28x%29) this is equivalent to 
%20%3D%20%5Cfrac%7B%5Csin(y)%7D%7B%5Ccos(y)%7D#card=math&code=x%20%3D%5Ctan%28y%29%20%3D%20%5Cfrac%7B%5Csin%28y%29%7D%7B%5Ccos%28y%29%7D)
Q: What are the domain and range for the tangent function?
A: The domain of 
#card=math&code=%5Ctan%28y%29) is given by the requirement that 
%20%5Cneq%200#card=math&code=%5Ccos%28y%29%20%5Cneq%200). When is %3D0#card=math&code=%5Ccos%28y%29%3D0)? Recall that
$\cos (y)=0 $ iff 
where 
is an integer. So the domain of the tangent funciton consists of all 
not of the form 
. (equivalently, not of the form 
)
The range of the tangent function is 
#card=math&code=%5Cmathbb%7BR%7D%3D%28-%5Cinfty%2C%20%2B%5Cinfty%29).
Thus, the domain of arctan(x) is 
, and the range of it is (under a standardlization) 
#card=math&code=%28-%5Cpi%2F2%2C%20%5Cpi%2F2%29).
Exercise: Determine the domain and range for the inverse cotangent function 
#card=math&code=y%20%3D%20%5Ccot%5E%7B-1%7D%28x%29).
Recall that 
%20%3D%20%5Cfrac%7B%5Ccos(y)%7D%7B%5Csin(y)%7D#card=math&code=%5Ccot%28y%29%20%3D%20%5Cfrac%7B%5Ccos%28y%29%7D%7B%5Csin%28y%29%7D) .
The inverse function theorem (on derivative)
Statement: If 
#card=math&code=y%3Dg%28x%29) is the inverse of 
#card=math&code=f%28x%29), then 
%7D%20%3D%20f’(X)%20%7C%7BX%20%3D%20g(x)%7D#card=math&code=%5Cfrac%7B1%7D%7Bg%27%28x%29%7D%20%3D%20f%27%28X%29%20%7C%7BX%20%3D%20g%28x%29%7D) .
Examples.
Let us set %3D%5Csin(x)#card=math&code=f%28x%29%3D%5Csin%28x%29) and its inverse 
%3D%5Carcsin(x)#card=math&code=g%28x%29%3D%5Carcsin%28x%29) .
%20%3D%20%5Ccos%20(X)#card=math&code=f%27%28X%29%20%3D%20%5Ccos%20%28X%29) . Now %7C%7BX%20%20%3D%20%5Carcsin(x)%7D%20%3D%20%5Ccos%20(%5Carcsin(x))#card=math&code=f%27%28X%29%7C%7BX%20%20%3D%20%5Carcsin%28x%29%7D%20%3D%20%5Ccos%20%28%5Carcsin%28x%29%29)
Denote 
#card=math&code=%5Ctheta%20%3D%20%5Carcsin%28x%29) . This means 
%20%3D%20x#card=math&code=%5Csin%28%5Ctheta%29%20%3D%20x) . And 
%3D#card=math&code=%5Ccos%28%5Ctheta%29%3D)?
By the Pythagorean theorem, 
%20%2B%20%5Ccos%5E2(%5Ctheta)%20%3D%201#card=math&code=%5Csin%5E2%28%5Ctheta%29%20%2B%20%5Ccos%5E2%28%5Ctheta%29%20%3D%201).
So %20%3D%20%5Ccos(%5Carcsin(x))%20%3D%20%5Cpm%20%5Csqrt%7B1-%5Csin%5E2(%5Ctheta)%7D%20%3D%20%5Cpm%20%5Csqrt%7B1-x%5E2%7D#card=math&code=%5Ccos%28%5Ctheta%29%20%3D%20%5Ccos%28%5Carcsin%28x%29%29%20%3D%20%5Cpm%20%5Csqrt%7B1-%5Csin%5E2%28%5Ctheta%29%7D%20%3D%20%5Cpm%20%5Csqrt%7B1-x%5E2%7D) .
Now using the knowledge on the range of 
and 
to determine the sign.
Recall that the range of 
#card=math&code=%5Ctheta%3D%5Carcsin%28x%29) is 
.So %20%3E0#card=math&code=%5Ccos%28%5Ctheta%29%20%3E0). Thus we choose the 
sign.
This implies that 
%20%3D%20(%5Carcsin(x))’%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1-x%5E2%7D%7D#card=math&code=g%27%28x%29%20%3D%20%28%5Carcsin%28x%29%29%27%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1-x%5E2%7D%7D).
Let us set 
%3D%5Ctan(x)#card=math&code=h%28x%29%3D%5Ctan%28x%29) and its derivative %3D%5Carctan(x)#card=math&code=g%28x%29%3D%5Carctan%28x%29).
%20%3D%5Csec%5E2(X)%20%3D%20%5Cfrac%7B1%7D%7B%5Ccos%5E2(X)%7D#card=math&code=h%27%28X%29%20%3D%5Csec%5E2%28X%29%20%3D%20%5Cfrac%7B1%7D%7B%5Ccos%5E2%28X%29%7D). Now replace 
by %20%3D%20%5Carctan(x)#card=math&code=g%28x%29%20%3D%20%5Carctan%28x%29), we get
)%20%20%3D%20%5Cfrac%7B1%7D%7B%5Ccos%5E2%20(%20%5Calpha)%7D#card=math&code=h%27%28%5Carctan%28x%29%29%20%20%3D%20%5Cfrac%7B1%7D%7B%5Ccos%5E2%20%28%20%5Calpha%29%7D) where 
#card=math&code=%5Calpha%20%3D%20%5Carctan%28x%29). (that is, 
%20%3D%20%5Cfrac%7B%5Csin(%5Calpha)%7D%7B%5Ccos(%5Calpha)%7D#card=math&code=x%3D%20%5Ctan%28%5Calpha%29%20%3D%20%5Cfrac%7B%5Csin%28%5Calpha%29%7D%7B%5Ccos%28%5Calpha%29%7D)) .
Squaring both sides, we get 
%7D%7B%5Ccos%5E2(%5Calpha)%7D%20%3D%20%5Cfrac%7B1-%5Ccos%5E2(%5Calpha)%7D%7B%5Ccos%5E2(%5Calpha)%7D#card=math&code=x%5E2%20%3D%20%5Cfrac%7B%5Csin%5E2%28%5Calpha%29%7D%7B%5Ccos%5E2%28%5Calpha%29%7D%20%3D%20%5Cfrac%7B1-%5Ccos%5E2%28%5Calpha%29%7D%7B%5Ccos%5E2%28%5Calpha%29%7D) This is equivalent to 
%7D%20%3D%201%2Bx%5E2#card=math&code=%5Cfrac%7B1%7D%7B%5Ccos%5E2%28%5Calpha%29%7D%20%3D%201%2Bx%5E2).
So )%20%3D%201%2Bx%5E2#card=math&code=h%27%28%5Carctan%28x%29%29%20%3D%201%2Bx%5E2). Thus, 
)’%20%3D%20%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D#card=math&code=%28%5Carctan%28x%29%29%27%20%3D%20%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D).
%3D(x%2B2)%2Fx%20%3D%201%2B2%2Fx#card=math&code=y%3Dg%28x%29%3D%28x%2B2%29%2Fx%20%3D%201%2B2%2Fx) 
#card=math&code=x%20%3D%202%2F%28y-1%29)
Switch the role of 
to obtain the inverse function expression: 
#card=math&code=y%20%3D%202%2F%28x-1%29) (because our convention to denote a function is using as variable) .
Now drop off the : the inverse function to %3D(x%2B2)%2Fx#card=math&code=g%28x%29%3D%28x%2B2%29%2Fx) is %20%3D%202%2F(x-1)#card=math&code=f%28x%29%20%3D%202%2F%28x-1%29)
%20%3D%202%2F(X-1)#card=math&code=f%28X%29%20%3D%202%2F%28X-1%29) and we find out %20%3D%202(-1)(X-1)%5E%7B-2%7D%20%3D%20-%5Cfrac%7B2%7D%7B(X-1)%5E2%7D#card=math&code=f%27%28X%29%20%3D%202%28-1%29%28X-1%29%5E%7B-2%7D%20%3D%20-%5Cfrac%7B2%7D%7B%28X-1%29%5E2%7D) .
Now, we replace by #card=math&code=g%28x%29) . So %7D%20%3D%20-%5Cfrac%7B2%7D%7B(1%2B%5Cfrac%7B2%7D%7Bx%7D%20-%201)%5E2%7D%20%3D%20-2(x%5E2%2F4)%3D-x%5E2%2F2#card=math&code=%5Cfrac%7B1%7D%7Bg%27%28x%29%7D%20%3D%20-%5Cfrac%7B2%7D%7B%281%2B%5Cfrac%7B2%7D%7Bx%7D%20-%201%29%5E2%7D%20%3D%20-2%28x%5E2%2F4%29%3D-x%5E2%2F2)
This is equivalent to say that %20%3D%20-2%2Fx%5E2#card=math&code=g%27%28x%29%20%3D%20-2%2Fx%5E2) .
Compute directly: %3D1%2B2%2Fx#card=math&code=g%28x%29%3D1%2B2%2Fx), so %20%3D%202(-1)x%5E%7B-2%7D%20%3D%20-2%2Fx%5E2#card=math&code=g%27%28x%29%20%3D%202%28-1%29x%5E%7B-2%7D%20%3D%20-2%2Fx%5E2) .
