Answer
$ f'(1)=24$
Equation of the tangent line:
$ y=24x-16$
Work Step by Step
Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Given that $ a=1$. Therefore, $ f'(1)=\lim\limits_{h \to 0}\frac{f(1+h)-f(1)}{h}=\lim\limits_{h \to 0}\frac{[8(1+h)^{3}]-(8\times1^{3})}{h}$
$=\lim\limits_{h \to 0}\frac{8(1^{3}+3h+3h^{2}+h^{3}-1)}{h}=\lim\limits_{h \to 0}\frac{8h^{3}+24h^{2}+24h}{h}=\lim\limits_{h \to 0}(8h^{2}+24h+24)=24$
Equation of the tangent line is of the form
$ y-f(a)=f'(a)(x-a)$
Knowing that $ f(a)=f(1)=8, f'(a)=f'(1)=24$ and $ a=1$, we obtain the equation of the tangent line through $ a $ as below.
$ y-8=24(x-1)$
Or $ y=24x-16$
