Differentiation of $\arccos v$

Let^5.10 $y = \arccos\ v$; then $y = \cos\ y$.

Figure: A single branch of the function $f(x)=\arccos(x)$.

Differentiating with respect to $y$ by XII, $\frac{dv}{dy} = -\sin\ y$, therefore, $\frac{dy}{dv} = -\frac{1}{\sin y}$, by (5.2). But since $v$ is a function of $x$, this may be substituted in the formula $\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}$, by (5.1), giving

$$\frac{dy}{dx} = -\frac{1}{\sin y} \cdot \frac{dv}{dx} = - \frac{1}{\sqrt{1 - v^2}} \frac{dv}{dx}$$

( $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - v^2}$, the plus sign of the radical being taken, since $\sin\, y$ is positive for all values of $y$ between 0 and $\pi$ inclusive). Therefore,

$$\frac{d}{dx}(\arccos\, v) = -\frac{\frac{dv}{dx}}{\sqrt{1 - v^2}}.$$

(equation (XIX) in §5.1 above).

Here's how to use SAGE to compute an example of this rule:

[fontsize=\scriptsize,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: t = var("t")
sage: x = var("x")
sage: solve(x == cos(t),t)
[t == acos(x)]
sage: f = solve(x == cos(t),t)[0].rhs()
sage: f
acos(x)
sage: diff(f,x)
-1/sqrt(1 - x^2)

This (1) computes $\arccos$ directly as the inverse function of $\cos$ (SAGE can use the notation ${\rm acos}$ in addition to $\arccos$), (2) computes its derivative.

Figure: The inverse cosine $\cos^{-1}\ x$ using SAGE.