SOLUTION 12: Begin with the function
$$f(x)= \arcsin x$$
and choose
$$x-values: 0.5 \rightarrow 0.45$$
so that
$$\Delta x = 0.45-0.5 = -0.05$$
The derivative of $\ y=f(x) \$ is
$$f'(x)= \displaystyle{ 1 \over \sqrt{1-x^2} }$$
The exact change of $y-$values is
$$\Delta y = f(0.45) - f(0.5)$$ $$= \arcsin(0.45) - \arcsin(0.5)$$ $$= \arcsin(0.45) - \displaystyle{ \pi \over 6 }$$ $$\approx \arcsin(0.45) - 0.5236$$
The Differential is
$$dy = f'(0.5) \ \Delta x$$ $$= \displaystyle{ 1 \over \sqrt{1-(0.5)^2} } \cdot (-0.05)$$ $$= \displaystyle{ 1 \over \sqrt{1-(1/2)^2} } \cdot (-0.05)$$ $$= \displaystyle{ 1 \over \sqrt{1-(1/4)} } \cdot (-0.05)$$ $$= \displaystyle{ 1 \over \sqrt{3/4} } \cdot (-0.05)$$ $$= \displaystyle{ 1 \over \sqrt{3}/2 } \cdot (-0.05)$$ $$= \displaystyle{ 2 \over \sqrt{3} } \cdot (-0.05)$$ $$= \displaystyle{ -0.1 \over \sqrt{3} }$$ $$\approx \displaystyle{ -0.1 \over 1.7321 }$$ $$\approx -0.0577$$
We will assume that
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\arcsin(0.45) - 0.5236 \approx -0.0577 \ \ \ \ \longrightarrow$$ $$\arcsin(0.45) \approx 0.5236 -0.0577 \ \ \ \ \longrightarrow$$ $$\arcsin(0.45) \approx 0.4659$$

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: $\arcsin(0.45) \approx 0.4668$