Differentiate 6x^2+2x+1 by first principles, showing every step in the process.
f(x) = 6x2+2x+1, f'(x)= [f(x+h)-f(x)]/h here is the original equation and the formula used to differentiate from first principles. For this proof the limit of h is: h=0 and should be stated throughout, but is not due to formatting problems.f'(x)=[6(x+h)2+2(x+h)+1 - (6x2+2x+1)]/h = [6(x2+2xh+h2)+2x+2h+1-6x2-2x-1]/h = [6x2+12xh+6h2+2x+2h+1-6x2-2x-1]/h here the formula are combined and brackets expanded.f'(x) =[12xh+6h2+2h]/h = 12x+6h+2, h=0 therefore 12x+6h+2= 12x+2, therefore f'(x) = 12x+2 the negatives from the previous line are resolved and the equation is canceled down to the answer.
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