SOLUTION 11: Begin with the function
$$ f(x)= \arctan x $$
and choose
$$ x-values: 1 \rightarrow 1.1 $$
so that
$$ \Delta x = 1.1-1 = 0.1 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ 1 \over 1+x^2 } $$
The exact change of $y-$values is
$$ \Delta y = f(1.1) - f(1) $$
$$ = \arctan(1.1) - \arctan(1) $$
$$ = \arctan(1.1) - \displaystyle{ \pi \over 4 } $$
$$ \approx \arctan(1.1) - 0.7854 $$
The Differential is
$$ dy = f'(1) \ \Delta x $$
$$ = \displaystyle{ 1 \over 1+(1)^2 } \cdot (0.1) $$
$$ = \displaystyle{ 1 \over 2 } (0.1) $$
$$ = 0.05 $$
We will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \arctan(1.1) - \displaystyle{ \pi \over 4 } \approx 0.05 \ \ \ \ \longrightarrow $$
$$ \arctan(1.1) \approx \displaystyle{ \pi \over 4 } + 0.05 \ \ \ \ \longrightarrow $$
$$ \arctan(1.1) \approx 0.7854 + 0.05 \ \ \ \ \longrightarrow $$
$$ \arctan(1.1) \approx 0.8354 $$
NOTE: The number 1 was chosen for its proximity to 1.1 and for it's convenient inverse tangent. Check the accuracy of the final estimate using a CALCULATOR: $ \arctan(1.1) \approx 0.8330 $
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