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: $\tan(\displaystyle{ \pi \over 4 }+0.15 ) \approx 1.3561$