## Calculus: Early Transcendentals 8th Edition

$f(x)=x^6$ and $a=2$
According to definition, $$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$$ Therefore, at a number a, the derivative would be $$f'(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Now we look at the given formula: $$f'(a)=\lim\limits_{h\to0}\frac{(2+h)^6-64}{h}$$ $$f'(a)=\lim\limits_{h\to0}\frac{(2+h)^6-2^6}{h}$$ Here, from the definition, we can deduce that $a=2$. That means $f(a)=2^6$. Therefore, $f(x)=x^6$