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 $
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