Differentiation Please can someone help me step by step with this sum Find dy/dx from first principles for 1) y=3x^2 + 1?

y = f(x) = 3x² + 1

f(x + h)
= 3(x + h)² + 1
= 3(x² + 2hx + h²) + 1
= 3x² + 6hx + 3h² + 1

[f(x + h) - f(x)] / h
= [(3x² + 6hx + 3h² + 1) - (3x² + 1)] / h
= [3x² + 6hx + 3h² + 1 - 3x² - 1] / h
= [6hx + 3h²] / h
= 6x + 3h

dy/dx = (d/dx)f(x)
= lim(h→0) {[f(x + h) - f(x)] / h}
= lim(h→0) (6x + 3h)
= lim(h→0)(6x) + lim(h→0)(3h)
= 6x

f ( x + h ) = 3 [ x + h ]² + 1
f ( x + h ) = 3 x² + 6 hx + 3h² + 1

[ f ( x + h ) - f (x) ] / h = [ 6 hx + 3h² ] / h = 6x + 3h

f ` (x) = 6x

f(x) = 3x^2+1
f(x+h)=3(x+h)^2 + 1
f(x+h)= 3(x^2+2xh+h^2) + 1
f(x+h) = 3x^2+6xh+3h^2+1
f(x+h)-f(x) = 3x^2+6xh+3h^2+1 - (3x^2+1)
f(x+h)-f(x) = 3x^2+6xh+3h^2+1 -3x^2-1
f(x+h)-f(x) = 6xh+3h^2
( f(x+h)-f(x)) / h = 6x + 3h

lim h-->0 ( f(x+h)-f(x)) / h = lim h-->0 (6x + 3h)
= 6x

dy/dx = 6x

lim(h-->0)[3(x + h)² + 1 - (3x² + 1)]/h =

lim(h-->0)(3x² + 6xh + 3h² + 1 - 3x² - 1)/h =

lim(h-->0)(6xh + 3h²)/h =

lim(h-->0)(6x + 3h) = 6x <==

This one is very easy
d(X^n)=n*X^(n-1)
d(C), where C is a number dC=0,
So d(y) yields d(3x^2+1)=d(3*x^2)+d(1)=3*2*x^(2-1)+0=6*x

　
y = f(x) = 3x² + 1

dy/dx = lim[h→0] (f(x+h) − f(x)) / h
dy/dx = lim[h→0] (3(x+h)² + 1 − (3x² + 1)) / h
dy/dx = lim[h→0] (3x² + 6hx + 3h² + 1 − 3x² − 1) / h
dy/dx = lim[h→0] (6hx + 3h²) / h
dy/dx = lim[h→0] h (6x + 3h) / h
dy/dx = lim[h→0] (6x + 3h)
dy/dx = 6x