Answer
$$\eqalign{
& \left( a \right).{\text{ }}m = 16 \cr
& \left( b \right).{\text{ }}y = 16x - 10 \cr} $$
Work Step by Step
$$\eqalign{
& \left( x \right) = 5{x^3} + x;{\text{ }}P\left( {1,6} \right) \cr
& \left( a \right){\text{ Calculate }}f'\left( x \right){\text{ using the definition of the derivative}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5{{\left( {x + h} \right)}^3} + \left( {x + h} \right) - \left( {5{x^3} + x} \right)}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5\left( {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right) + x + h - 5{x^3} - x}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5{x^3} + 15{x^2}h + 15x{h^2} + 5{h^3} + h - 5{x^3}}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{15{x^2}h + 15x{h^2} + 5{h^3} + h}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {15{x^2} + 15xh + 5{h^2} + 1} \right) \cr
& f'\left( x \right) = 15{x^2} + 15x\left( 0 \right) + 5{\left( 0 \right)^2} + 1 \cr
& f'\left( x \right) = 15{x^2} + 1 \cr
& {\text{Calculate the slope at }}P\left( {1,6} \right) \cr
& m = f'\left( 2 \right) = 15{\left( 1 \right)^2} + 1 = 16 \cr
& \cr
& \left( b \right){\text{ Find an equation of the tangent line at }}P\left( {1,6} \right) \cr
& y - 6 = 16\left( {x - 1} \right) \cr
& y - 6 = 16x - 16 \cr
& {\text{ }}y = 16x - 10 \cr} $$
