Answer
$${\text{ }}{\left( {1 + x} \right)^{15}} \approx 1 + 15x$$
Work Step by Step
$$\eqalign{
& {\left( {1 + x} \right)^{15}} \approx 1 + 15x,{\text{ }}{x_0} = 0 \cr
& {\text{Let }}f\left( x \right) = {\left( {1 + x} \right)^{15}} \cr
& {\text{ }}f\left( {{x_0}} \right) = f\left( 0 \right) = {\left( {1 + 0} \right)^{15}} = 1 \cr
& f'\left( x \right) = {\left( {1 + x} \right)^{15}} = 15{\left( {1 + x} \right)^{14}} \cr
& f'\left( {{x_0}} \right) = f'\left( 0 \right) = 15{\left( {1 + 0} \right)^{14}} = 15 \cr
& \cr
& {\text{We can approximate values of }}f\left( x \right){\text{ by}} \cr
& {\text{ }}f\left( x \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \cr
& {\text{then}} \cr
& {\text{ }}{\left( {1 + x} \right)^{15}} \approx 1 + 15\left( {x - 0} \right) \cr
& {\text{ }}{\left( {1 + x} \right)^{15}} \approx 1 + 15x \cr} $$
