Help with derivatives?

let f(x) = 6x^2 +x. Use the definition of the derivative as the limit of the difference quoetient to find f '(x).

$\begin{array}{rcll} f(x) &=& 6x^2+x\\ f(x+h)&=&6(x+h)^2+(x+h)\\\\ \text{The difference quotient is:}\\ \dfrac{\Delta x}{\Delta y} &=& \dfrac{f(x+h)-f(x)}{h}\\ &=& \dfrac{6(x+h)^2+(x+h)-(6x^2+x)}{h}\\ &=& \dfrac{6(x+h)^2+ x +h-6x^2 - x }{h}\\ &=& \dfrac{6(x+h)^2 +h-6x^2 }{h}\\ &=& \dfrac{6(x^2+2xh+h^2) +h-6x^2 }{h}\\ &=& \dfrac{6x^2+12xh+6h^2 +h-6x^2 }{h}\\ &=& \dfrac{ 12xh+6h^2 +h }{h}\\ &=& \dfrac{ h( 12x+6h +1) }{h}\\ \dfrac{\Delta x}{\Delta y} &=& 12x+6h +1 \end{array}$

$\begin{array}{rcll} f'(x) &=& \dfrac{dy}{dx} \\ \dfrac{dy}{dx} &=& \lim \limits_{h\to 0} { \left( 12x+6h +1 \right) } \\ \dfrac{dy}{dx} &=& \left( 12x+6\cdot 0 +1 \right) \\ \dfrac{dy}{dx} &=& \left( 12x+1 \right) \\\\ \mathbf{f'(x)} &\mathbf{=}& \mathbf{12x+1} \end{array}$

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(x+6 x^2)
The derivative of f(x) is f'(x):
f'(x) = d/dx(x+6 x^2)
Differentiate the sum term by term and factor out constants:
f'(x) = d/dx(x)+6 d/dx(x^2)
The derivative of x is 1:
f'(x) = 6 (d/dx(x^2))+1
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
f'(x) = 1+6 2 x
Simplify the expression:
f'(x) = 1+12 x
Expand the left hand side:
Answer: |  f'(x) = 1+12x