Question

I had the function f(x)=arccos(x^2+1) and received the antiderivative -2x/√(1-(x^2+1)^2).

Can you help me form a function table for f(x)=arccos(x^2+1) and -2x/√(1-(x^2+1)^2) to show when they`re defined?

Answer 1

I need to prove when they´re defined somehow, but I have no idea. Maybe a function table is the wrong way to go?

Answer 2

arocos(u) is defined for u between -1 and 1

Answer 3

\[
x^2+1 \ge 1\]

Answer 4

So you only can take x=0

Answer 5

\[arccos (x^2+1)= \alpha\\cos\alpha=x^2 +1\]
Take derivative both sides
\[-sin \alpha \dfrac{d}{dx}\alpha =2x\\\dfrac{d}{dx}\alpha =\dfrac{-2x}{sin\alpha}\]

Now, back to the first line, \(\alpha '= (arccos (x^2+1))'\)

So, \((arccos(x^2+1))'=\dfrac{-2x}{sin\alpha}\)

Now, \(cos^2\alpha +sin^2\alpha =1\\sin\alpha =\sqrt{1-cos^2\alpha}\)

Hence \((arccos\alpha)'=\dfrac{-2x}{\sqrt{1-(x^2+1)^2}}\)

Answer 6

Thank you so much for the help!

Answer 7

since
\[
0\le \cos(\alpha)=x^2 +1\le 1\implies x=0

\]
So the function is constant and its derivative is zero

Answer 8

the function f(x)