Proof Correction: $\cos ' (x) = -2 \sin (x)$ Preliminary Information
Let $$A(y) = \cfrac{y \sqrt{1 - y^2}}{2} + \int_{y}^{1} 1\sqrt{1 - t^2} dt \ \ \ \text{on $[-1, 1]$}$$ 
Moreover $$ A'(y) = \dfrac{-1}{2\sqrt{1 - y^2}}$$
Define $\cos x$ as $A(\cos x) = \cfrac{x}{2}$ and $\sin x = \sqrt{1 - \cos ^{2} x}$
Problem: Find $\cos' (x)$. 
$A(\cos x) = \dfrac{x}{2}$; since $A$ is decreasing it is one-one and $A^{-1}$ a function. Consider that $A^{-1}\left( \dfrac{x}{2} \right) = \cos x$. Let us find then $(A^{-1})' \left (\dfrac{x}{2} \right)$.
$\begin{equation}
\begin{split}
(A^{-1})' \left (\dfrac{x}{2} \right) &= \dfrac{1}{A'(A^{-1} \left (\dfrac{x}{2} \right)} \\
&= -2 \sin x
\end{split}
\end{equation}$
Textbook's Proof


You forgot to apply the chain rule and multiply by $(x/2)'$ when differentiating $(A^{-1})(x/2)$.
