Answer
$$f\left( {x + \Delta x} \right) \approx 0 + 1\left( {0.05} \right)$$
Work Step by Step
$$\eqalign{
& \tan 0.05 \cr
& {\text{Using }}f\left( x \right) = \tan x \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\tan x} \right] \cr
& f'\left( x \right) = {\sec ^2}x \cr
& {\text{Using the formula}} \cr
& f\left( {x + \Delta x} \right) \approx f\left( x \right) + f'\left( x \right)dx \cr
& {\text{We can write,}} \cr
& f\left( {x + \Delta x} \right) \approx \tan x + {\sec ^2}xdx \cr
& f\left( {x + \Delta x} \right) = \tan 0.05 \cr
& {\text{Now, choosing }}x = 0{\text{ and }}dx = 0.05,{\text{ and substituting }} \cr
& f\left( {x + \Delta x} \right) \approx \tan \left( 0 \right) + {\sec ^2}\left( 0 \right)\left( {0.05} \right) \cr
& f\left( {x + \Delta x} \right) \approx 0 + 1\left( {0.05} \right) \cr} $$
