Answer
$ f'(4)=100$
Equation of the tangent line:
$ y=100t-256$
Work Step by Step
Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Given that $ a=4$. Therefore, $ f'(4)=\lim\limits_{h \to 0}\frac{f(4+h)-f(4)}{h}=\lim\limits_{h \to 0}\frac{[2(4+h)^{3}+4(4+h)]-(2\times4^{3}+4\times4)}{h}$
$=\lim\limits_{h \to 0}\frac{[2(4^{3}+3\times4^{2}\times h+3\times4\times h^{2}+h^{3})+16+4h]-144}{h}=\lim\limits_{h \to 0}\frac{2h^{3}+24h^{2}+100h}{h}=\lim\limits_{h \to 0}(2h^{2}+24h+100)=100$
Equation of the tangent line is of the form
$ y-f(a)=f'(a)(t-a)$
Knowing that $ f(a)=f(4)=144, f'(a)=f'(4)=100$ and $ a=4$, we obtain the equation of the tangent line through $ a $ as below.
$ y-144=100(t-4)$
Or $ y=100t-256$
