Explanation

Substitute x by (x + h) to have f(x + h) = 3(x + h)\(^2\) + 2x + 2h - 1 - 3x\(^2\) - 2x + 1

= 6\(x\)h + 3h\(^2\) + 2h

Divided through by h and obtained \(\frac{f(x + h) - f(x)}{h} = \frac{6 \times h + 3h^2 + 2h}{h} = 6x + 3h + 2\)

Taking limit as h tends to zero, lim\(_{h \to o}\) \(\frac{f(x+h) -f(x)}{h}\) 

= lim\(_{h \to o}\) (6x + 3h + 2) so that f\(^1\)(x) = 6x + 2

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