Solution:
f(x)=3x2β4x+7
By definition of derivatives:
fβ²(x)β=hβ0limβhf(x+h)βf(x)β=hβ0limβh(3(x+h)2β4(x+h)+7)β(3x2β4xβ7)β=hβ0limβh(3(x2+2xh+h2)β4xβ4h)β(3x2β4x)β=hβ0limβh3x2+6xh+3h2β4xβ4hβ3x2+4xβ=hβ0limβh6xh+3h2β4hβ=hβ0limβ(6x+3hβ4)=6xβ4β
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