The derivative of $\arcsin(x)$ at $x=1$ Why the derivative of $\arcsin(x)$ isn't defined at $x=1$? The functional value of $\arcsin(x)$ seems to grow faster around $1$, but there is no discontinuity. So, why the slope is undefined?
 A: The tangent line becomes vertical as $x$ approaches $1^-$, and vertical lines have undefined slope.  This corresponds to $\sin x$ having a horizontal tangent (slope $0$) at $\pi/2$.
A: Take
$$y=\arcsin x\implies x=\sin y$$
$y$ is an angle in a right triangle and $\sin y=\frac{\text{opp}}{\text{hyp}}=x$. Therefore, we can take the opposite side to $y$ to be $x$ and the hypotenuse to be $1$, making the adjacent side $\sqrt{1-x^2}$.
$$\frac{d}{dx}x=\frac{d}{dx}\sin y\implies\frac{dy}{dx}=\sec y=\frac{1}{\sqrt{1-x^2}}$$
Which is undefined at $x=\pm1$
A: Geometrically, arcsine is the reflection of sine about the line $y = x$: the "inverse function" operation exchanges the roles of the independent and dependent variable. This means that its graph is effectively a sine curve along the $y$-axis, but of course that's now a one-to-many relation, not a function (since sine is many-to-one), so we have to limit it and choose a small piece.  This piece is conventionally taken from $y = -\pi/2$ to $y = +\pi/2$, which are crests of the sine wave. The horizontal tangents at the crests become vertical tangents in the mirror images of them, and thus their slope is undefined.
Alternatively, you can think in a somewhat loose sense that the derivative $y$ as a function of $x$ is $\frac{dy}{dx}$, but for $x$ as a "function" of $y$ it is "$\frac{dx}{dy}$", which is the reciprocal, and thus wherever the derivative of sine is 0, the corresponding point on the inverse sine will have undefined (reciprocal of 0) derivative.
