How do you find f'(x) using the definition of a derivative for #(sqrt2x) - x^3#?

Question

1562 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-06T11:06:07

#f'(x)=1/sqrt(2x)-3x^2#

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

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

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

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

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

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

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

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

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

#f'(x)=2/(2sqrt(2x)) - 3x^2=1/sqrt(2x)-3x^2#