Differentiability of arcsin(x) at the domain endpoint x = 1

The difference quotient at $a=1$ is $\dfrac{\arcsin(x) - \pi/2}{x-1}$, and since the domain of $\arcsin$ is $[-1,1]$, only $x \to 1^-$ can be considered. As $x \to 1^-$, the numerator $\arcsin(x) - \pi/2 \to 0^-$ while the denominator $x - 1 \to 0^-$, giving a quotient of the form $\frac{0^-}{0^-}$, which is indeterminate.

To analyze this rigorously, we use the known derivative formula: $\dfrac{d}{dx}[\arcsin(x)] = \dfrac{1}{\sqrt{1-x^2}}$. As $x \to 1^-$, we have $1 - x^2 \to 0^+$, so $\sqrt{1-x^2} \to 0^+$, and therefore $\dfrac{1}{\sqrt{1-x^2}} \to +\infty$. This means $\lim_{x\to 1^-} \dfrac{\arcsin(x)-\pi/2}{x-1} = +\infty$. Since the limit does not exist as a finite number, $f'(1)$ does not exist. Graphically, this corresponds to the tangent line at $x=1$ being vertical, which is a classic case where a function fails to be differentiable despite being continuous there.

At $a = -1$, the difference quotient is $\dfrac{\arcsin(x) - (-\pi/2)}{x-(-1)} = \dfrac{\arcsin(x)+\pi/2}{x+1}$, and only $x \to -1^+$ is meaningful since the domain excludes $x < -1$. As $x \to -1^+$, we have numerator $\arcsin(x) + \pi/2 \to 0^+$ (since $\arcsin(x) \to -\pi/2$ from above) and denominator $x + 1 \to 0^+$, giving an indeterminate form $\frac{0^+}{0^+}$.

Using the derivative formula $\dfrac{d}{dx}[\arcsin(x)] = \dfrac{1}{\sqrt{1-x^2}}$, as $x \to -1^+$ we have $1 - x^2 \to 0^+$, so $\sqrt{1-x^2} \to 0^+$, and therefore $\dfrac{1}{\sqrt{1-x^2}} \to +\infty$. Thus $\lim_{x \to -1^+} \dfrac{\arcsin(x)+\pi/2}{x+1} = +\infty$, so the one-sided limit is infinite and $f'(-1)$ does not exist. This matches the graph, which shows a vertical tangent line at $x=-1$ just as it did at $x=1$, so $f$ is not differentiable at either endpoint of its domain.

The tangent line is vertical at both x equals one and x equals negative one. This happened because the difference quotient grew without bound at each endpoint, meaning the slope approached infinity rather than a finite value. A vertical tangent line is a classic graphical signal that a function is not differentiable at that point, even though the function remains continuous there.