SOLUTION 9: Begin with the function
$$ f(x)= \tan x $$
and choose
$$ x-values: \displaystyle{ \pi \over 4 } \rightarrow \displaystyle{ \pi \over 4 } + 0.15 $$
so that
$$ \Delta x = (\displaystyle{ \pi \over 4 } + 0.15) - \displaystyle{ \pi \over 4 } = 0.15 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \sec^2 x $$
The exact change of $y-$values is
$$ \Delta y = f(\displaystyle{ \pi \over 4 } + 0.15) - f(\displaystyle{ \pi \over 4 }) $$
$$ = \tan(\displaystyle{ \pi \over 4 } + 0.15) - \tan(\displaystyle{ \pi \over 4 }) $$
$$ = \tan(\displaystyle{ \pi \over 4 } + 0.15) - 1 $$
The Differential is
$$ dy = f'(\displaystyle{ \pi \over 4 }) \ \Delta x $$
$$ = \sec^2(\displaystyle{ \pi \over 4 }) \cdot (0.15) $$
$$ = \displaystyle{ 1 \over \cos^2(\displaystyle{ \pi \over 4 }) } \cdot (0.15) $$
$$ = \displaystyle{ 1 \over \Big( \cos(\displaystyle{ \pi \over 4 })\Big)^2 } \cdot (0.15) $$
$$ = \displaystyle{ 1 \over \Big( { \sqrt{2} \over 2 } \Big)^2 } \cdot (0.15) $$
$$ = \displaystyle{ 1 \over { 2 \over 4 } } \cdot (0.15) $$
$$ = \displaystyle{ 1 \over { 1 \over 2 } } \cdot (0.15) $$
$$ = (2) (0.15) $$
$$ = 0.3 $$
We will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \tan(\displaystyle{ \pi \over 4 } + 0.15) - 1 \approx 0.3 \ \ \ \ \longrightarrow $$
$$ \tan(\displaystyle{ \pi \over 4 } + 0.15) \approx 1+0.3 \ \ \ \ \longrightarrow $$
$$ \tan(\displaystyle{ \pi \over 4 } + 0.15) \approx 1.3 \ \ \ \ \longrightarrow $$
NOTE: The number $ \displaystyle{ \pi \over 4 } $ was chosen for its proximity to $ \displaystyle{ \pi \over 4 }+0.15 $ and for it's convenient tangent value. Check the accuracy of the final estimate using a CALCULATOR: $ \tan(\displaystyle{ \pi \over 4 }+0.15 ) \approx 1.3561 $
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