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.

TN
Answered by Thomas N. Maths tutor

6470 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The equation kx^2 + 4x + (5 – k) = 0, where k is a constant, has 2 different real solutions for x. Show that k satisfies k^2-5k+4>0.


integrate from 0 to 2: 2x*sqrt(x+2) dx


Find the area under the curve y = sin(2x) + cos(x) between 0 and pi/2


Integrate (x^2 +2)(2x-6) with respect to x.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning