Answer
$$f'(a)=6a^2+1$$
Work Step by Step
$$f(t)=2t^3+t$$
That means $$f(a)=2a^3+a$$
According to definition, the derivative of $f(t)$ at number $a$ is $$f'(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$
$$f'(a)=\lim\limits_{h\to0}\frac{[2(a+h)^3+(a+h)]-[2a^3+a]}{h}$$
$$f'(a)=\lim\limits_{h\to0}\frac{[(2a^3+2h^3+6a^2h+6ah^2)+(a+h)]-2a^3-a}{h}$$
$$f'(a)=\lim\limits_{h\to0}\frac{2h^3+6a^2h+6ah^2+h}{h}$$
$$f'(a)=\lim\limits_{h\to0}(2h^2+6ah+6a^2+1)$$
$$f'(a)=2\times0^2+6a\times0+6a^2+1$$
$$f'(a)=6a^2+1$$
