Question

How do I find the derivative of `x^3 - 2x^2 + x/4 +6` using first principles?

Answer 1

First Principles `->` Difference Quotient

`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`

`f(x)=x^3-2x^2+x/4+6`

`f(x+h)=(x+h)^3-2(x+h)^2+(x+h)/4+6`

`f'(x)=lim_(h->0)((x+h)^3-2(x+h)^2+(x+h)/4+6-(x^3-2x^2+x/4+6))/h`

`f'(x)=lim_(h->0)((x+h)^3-2(x+h)^2+(x+h)/4+6-x^3+2x^2-x/4-6)/h`

`f'(x)=lim_(h->0)((x+h)^3-2(x+h)^2+(x+h)/4-x^3+2x^2-x/4)/h`

`f'(x)=lim_(h->0)((x+h)^3-2(x^2+2xh+h^2)+(x+h)/4-x^3+2x^2-x/4)/h`

`f'(x)=lim_(h->0)((x+h)^3-2x^2-4xh-2h^2+(x+h)/4-x^3+2x^2-x/4)/h`

`f'(x)=lim_(h->0)((x+h)^3-4xh-2h^2+(x+h)/4-x^3-x/4)/h`

`f'(x)=lim_(h->0)((x+h)(x^2+2xh+h^2)-4xh-2h^2+x/4+h/4-x^3-x/4)/h`

`f'(x)=lim_(h->0)((x+h)(x^2+2xh+h^2)-4xh-2h^2+h/4-x^3)/h`

`f'(x)=lim_(h->0)(x^3+2x^2h+xh^2+hx^2+2xh^2+h^3-4xh-2h^2+h/4-x^3)/h`

`f'(x)=lim_(h->0)(2x^2h+xh^2+hx^2+2xh^2+h^3-4xh-2h^2+h/4)/h`

`f'(x)=lim_(h->0)(h*(2x^2+xh+x^2+2xh+h^2-4x-2h+1/4))/h`

`f'(x)=lim_(h->0)2x^2+xh+x^2+2xh+h^2-4x-2h+1/4`

`f'(x)=2x^2+x(0)+x^2+2x(0)+(0)^2-4x-2(0)+1/4`

`f'(x)=2x^2+x^2-4x+1/4`

`f'(x)=3x^2-4x+1/4`