SOLUTION 13: Begin with the function
$$f(x)= \arcsin x$$
and choose
$$x-values: 0 \rightarrow 0.12$$
so that
$$\Delta x = 0.12-0 = 0.12$$
The derivative of $\ y=f(x) \$ is
$$f'(x)= \displaystyle{ 1 \over \sqrt{1-x^2} }$$
The exact change of $y-$values is
$$\Delta y = f(0.12) - f(0)$$ $$= \arcsin(0.12) - \arcsin(0)$$ $$= \arcsin(0.12) - 0$$ $$= \arcsin(0.12)$$
The Differential is
$$dy = f'(0) \ \Delta x$$ $$= \displaystyle{ 1 \over \sqrt{1-(0)^2} } \cdot (0.12)$$ $$= \displaystyle{ 1 \over \sqrt{1-0} } \cdot (0.12)$$ $$= \displaystyle{ 1 \over \sqrt{1} } \cdot (0.12)$$ $$= (1) (0.12)$$ $$= 0.12$$
We will assume that
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\arcsin(0.12) \approx 0.12$$

NOTE: The number 0.5 was chosen for its proximity to 0.45 and for it's convenient inverse tangent. Check the accuracy of the final estimate using a CALCULATOR: $\arcsin(0.12) \approx 0.1203$