Strange result

[Steven G]

I would have thought that if you compute the derivative of f(x) and -f(x) and got the same answer then f(x) would be a constant.

Here is a proof of that statement: f'(x) = -f'(x)=> 2f'(x) = 0 => f'(x) = 0 => f(x) = c, a constant.

However that does not seem to be true. It is not true for f(x)= cos^-1(x) and cos^-1(x) is not a constant.

To see what I am saying about f(x) = cos^-1(x):

Let y = cos^-1(x) and y lies in 1st quadrant.

Then cos (y) =x (in the right triangle assume the adj side of y is x, opposite is sqrt(1-x²) and hyp is 1)

Taking the derivative of both sides yields 1=-sin(y)y' or y' = -1/sin(y) = -1/sqrt(1-x²).

Now let y=-cos^-1(x) with y in 1st quadrant (or -y in quad IV).

Then cos (-y) =x (in the right triangle assume the adj side of y is x, opposite is -sqrt(1-x²) and hyp is 1)

Taking the derivative of both sides yields 1=sin(-y)y' or y' = 1/sin(-y) = -1/sqrt(1-x²).

What is going on here???

 

[pka]

However that does not seem to be true. It is not true for f(x)= cos^-1(x) and cos^-1(x) is not a constant.
To see what I am saying about f(x) = cos^-1(x):
Let y = cos^-1(x) and y lies in 1st quadrant.
Then cos (y) =x (in the right triangle assume the adj side of y is x, opposite is sqrt(1-x²) and hyp is 1)
Taking the derivative of both sides yields 1=-sin(y)y' or y' = -1/sin(y) = -1/sqrt(1-x²).
Now let y=-cos^-1(x) with y in 1st quadrant (or -y in quad IV).
Then cos (-y) =x (in the right triangle assume the adj side of y is x, opposite is -sqrt(1-x²) and hyp is 1)
Taking the derivative of both sides yields 1=sin(-y)y' or y' = 1/sin(-y) = -1/sqrt(1-x²).
What is going on here???

I think what going on is you miss-understanding the function \(\displaystyle arccos(x)\).
Its domain is \(\displaystyle [-1,1]\) and its range is \(\displaystyle [0,\pi]\). Look at its graph.
So that there is no y in the fourth quadrant to consider.
\(\displaystyle \frac{d}{dx}[\arccos(x)]=\dfrac{-1}{\sqrt{1-x^2}}\)

 

[Steven G]

I think what going on is you miss-understanding the function \(\displaystyle arccos(x)\).
Its domain is \(\displaystyle [-1,1]\) and its range is \(\displaystyle [0,\pi]\). Look at its graph.
So that there is no y in the fourth quadrant to consider.
\(\displaystyle \frac{d}{dx}[\arccos(x)]=\dfrac{-1}{\sqrt{1-x^2}}\)

Thanks for pointing that out to me, especially since I already knew it. Sorry for making you spend your time on it. I truly appreciate it.