I have to determine where the function $$ f:x \mapsto \arccos \frac{1}{\sqrt{1+x^2}} $$ is differentiable. For $ x = 0$ we have $\arccos (1) = 0$ so would f be differentiable in the interval from $[-1,1[$?
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Try taking the general derivative. The points where the derivative does not exist, are the points where it is non-derivable. – Ishraaq Parvez Feb 25 '20 at 13:27
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I think it should be not derivable at x = 0. – Ishraaq Parvez Feb 25 '20 at 13:29
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hat it is differentiable on the open interval $(-1,1)$.
Now, as $\frac 1{\sqrt{1+x^2}}$ is differentiable on $\mathbf R$, the condition would be $-1 < \frac 1{\sqrt{1+x^2}}< 1$, but as it is positive for all $x$, this condition reduces to $$\sqrt{1+x^2}\ne 1\iff x\ne 0.$$
Bernard
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