Answer
$a.$ The value of the derivative is $f'(10)=600.$
$b.$ The equation is $y=600x-4000.$
Work Step by Step
$a.$ By definition of a derivative at point $a=10$ we have
$$f'(10)=\lim_{h\to0}\frac{f(10+h)-f(10)}{h}=\lim_{h\to0}\frac{2\cdot(10+h)^3-2\cdot 10^3}{h}=\lim_{h\to0}\frac{2(10^3+3h10^2+3h^2\cdot 10+h^3)-2\cdot10^3}{h}=\lim_{h\to0}\frac{2\cdot10^3+6\cdot10^2h+60h^2+2h^3-2\cdot10^3}{h}=\lim_{h\to0}\frac{2h^3+600h+60h^2}{h}=\lim_{h\to0}(2h^2+600+60h)=2\cdot0^2+600+60\cdot0=600.$$
$b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=10$, $f(a)=2\cdot10^3=2000$ and $f'(a)=600$ we get
$$y-2000=600(x-10)\Rightarrow y-2000=600x-6000$$ which gives $$y=600x-4000.$$
