Question

find the derivative by first principle
f(x)= x^2-6÷3x

Answer 1

 solution

given that  y= x²-6/3x

     ⇒y+бy=(x+бx)²-6/3(x+бx)

      y+бy=(x²+2xбx+бx²-6)/(3x+3бx)

     by subtracting y from both side

   ⇒бy=(x²+2xбx+бx²-6)/(3x+3бx)-y   but y=x²-6/3x

 ⇒бy=(x²+2xбx+бx²-6)/(3x+3бx)-x²-6/3x

 ⇒бy={3(x²+2xбx+бx²-6)-(x²-6)(3x+бx)}/3x(3x+3бx)

  бy/бx=(3x³+6x²бx+3xбx²-18x)-(3x³+3x²бx-18x-18бx)/3x(3x+3бx)

  by taking the limiting value at бx⇒0

  бy/бx=3x²+3xбx+18бx/9x²

  ⇒dy/dx=(3x²+18)/9x²