The derivative of cos^-1(cosx); y' = ?

Question

I kind of know how to start it off:

The Inverse Trig. Function for cos^-1 is -1/ sq root (1-x^2).

So it would be [-1/ sq root(1-cos^2(x))] * d/dx cos(x). d/dx cos(x) is equal to -sin(x).

So then you would have sin(x)/ sq root(1-cos^2(x)).

First off, am I right. Second, isn't my answer an identity?

Answer 1

Yes and yes. But you should have 'almost' seen that from arccos(cos(c)) which reads, 'the angle whose cosine is cosine(x)' so the derivative would be 1.
And yes, sin(x) = √(1-cos(x)) since sin²(x) + cos²(x) = 1

Good job ☺


Doug

Answer 2

This is pathetic,
the arccos function(which is the same as cos^-1) is the inverse of the cos function.
therefore,
arccos (cos x) = x
and the derivative of x is 1.
Even if you do the unneccesary derivative dance you still get the same result as V(1 - cos^2 x ) = V(sin^2 x) = sin x

Answer 3

You can apply simpler approach

simplify the expression first

y = cos^-1(cos x) = x by definition (assuing x in 1st quadrant)

so dy/dx = 1

similarly you can do for

y = sin^-1( cosx ) = pi/4 -x

dy/dx = -1

this may not be possible in all cases but in case you recognize the case you can simplify and do it. This was your case

but of you have in general y = g^(-1) f(x)
convert to
g(y) = f(x)
and differetiating both sides you get

dg(y)/dy dy/dx = df/dx
dy/dx = df/dx . (1/dg(y)/dy)