0Differentiate 2x² using first principle
We know from first principle that dy/dx f(x) = lim δx→0 [ f( x + δx ) - f (x) ] /δx
Here f(x) = 2 x² substituting 2x² in place of x in the equation above
lim δx → 0 [ 2 ( x + δx )² - 2 x² ] /δx
lim δx → 0 [ 2 ( x² + δx² + 2 • x • δx ) - 2 x² ] / δx
lim δx → 0 [ 2 x² + 2 δx² + 4 x δx - 2x² ] / δx
lim δx → 0 [ 2 δx² + 4 x δx ] / δx. ( 2 x² cancels out )
lim δx → 0 ( 2 δx² ) / δx + lim δx → 0 ( 4 x δx ) / δx
lim δx → 0 ( 2 δx ) + lim δx → 0 ( 4 x ).
= 0 + 4 x = 4 x
∴ from first principle [ dy/dx ( 2 x² ) ] = ( 4 x )
