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 $
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