Let #f(x)=x+1/x^3#
#f(x)=(x^4+1)/x^3#
First principles tells us that:
#f'(x)=lim_(h->0)(f(x+h)-f(x))/h#
So, we have:
#f'(x)=lim_(h->0)(((x+h)^4+1)/(x+h)^3-(x^4+1)/x^3)/h#
#color(white)(f'(x))=lim_(h->0)((x^3((x+h)^4+1)-(x+h)^3(x^4+1))/(x(x+h))^3)/h#
#color(white)(f'(x))=lim_(h->0)(x^3((x+h)^4+1)-(x+h)^3(x^4+1))/(h(x^2+hx)^3)#
#color(white)(f'(x))=lim_(h->0)(x^3(h^4+4h^3x+6h^2x^2+4hx^3+x^4+1)-(x^3+3hx^2+3h^2x+h^3)(x^4+1))/(h(h^3x^3+3h^2x^4+3hx^5+x^6))#
#color(white)(f'(x))=lim_(h->0)(x^3h^4+4h^3x^4+6h^2x^5+4hx^6+x^7+x^3-x^7-3hx^6-3h^2x^5-h^3x^4-x^3-3hx^2-3h^2x-h^3)/(h^4x^3+3h^3x^4+3h^2x^5+hx^6)#
#color(white)(f'(x))=lim_(h->0)(h(x^6+3hx^5+3h^2x^4+h^3x^3-3x^2-3hx-h^2))/(h^4x^3+3h^3x^4+3h^2x^5+hx^6)#
#color(white)(f'(x))=lim_(h->0)(x^6+3hx^5+3h^2x^4+h^3x^3-3x^2-3hx-h^2)/(h^3x^3+3h^2x^4+3hx^5+x^6)#
#color(white)(f'(x))=(x^6+3(0)x^5+3(0)^2x^4+(0)^3x^3-3x^2-3(0)x-(0)^2)/((0)^3x^3+3(0)^2x^4+3(0)x^5+x^6)#
#color(white)(f'(x))=(x^6-3x^2)/(x^6)#
#color(white)(f'(x))=1-3/x^4#
#color(white)(f'(x))=1-3x^(-4)#
Proof:
#f(x)=x+x^(-3)#
#f'(x)=1+x^(-3-1)*(-3)=1-3x^(-4)#
