Answer
$f'(1)=3$
The equation of the tangent line is $$(l):y=3x-1$$
Work Step by Step
$$f(x)=3x^2-x^3$$
1) The derivative of $f(x)$ at a number $a$ is calculated as follows: $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$
$$f'(a)=\lim\limits_{x\to a}\frac{(3x^2-x^3)-(3a^2-a^3)}{x-a}$$
$$f'(a)=\lim\limits_{x\to a}\frac{3(x^2-a^2)-(x^3-a^3)}{x-a}$$
$$f'(a)=\lim\limits_{x\to a}\frac{3(x-a)(x+a)-(x-a)(x^2+a^2+ax)}{x-a}$$
$$f'(a)=\lim\limits_{x\to a}\frac{(x-a)[3(x+a)-x^2-a^2-ax]}{x-a}$$
$$f'(a)=\lim\limits_{x\to a}[3(x+a)-x^2-a^2-ax]$$
$$f'(a)=3(a+a)-a^2-a^2-a\times a$$
$$f'(a)=6a-3a^2$$
So $f'(1)=6\times1-3\times1^2=3$
2) The equation of the tangent line $l$ of the curve $y$ at point $(1,2)$ would have the following form $$(l):y=f'(1)x+b$$$$(l):y=3x+b$$
Since point $(1,2)$ lies in the tangent line $l$, we can find $b$ by applying the point into the equation of the tangent line $l$: $$3\times1+b=2$$$$3+b=2$$$$b=-1$$
Overall, the equation of tangent line $l$ is $$(l):y=3x-1$$
