How do you use the limit definition of the derivative to find the derivative of #f(x)=sqrt(x)-2x+1#?

1 Answer
Jan 16, 2018

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

Explanation:

The limit definition says that for #f(x)#, #f'(x)=lim_(h->0)(f(x+h)-f(x))/h#

For #f(x)=sqrt(x)-2x+1#, we have:
#f'(x)=lim_(h->0)(sqrt(x+h)-2(x+h)+1-sqrt(x)+2x-1)/h#

#color(white)(llllllll)=lim_(h->0)(sqrt(x+h)-2x-2h+1-sqrt(x)+2x-1)/h#

#color(white)(llllllll)=lim_(h->0)(sqrt(x+h)-2h-sqrt(x))/h#

#color(white)(llllllll)=lim_(h->0)(sqrt(x+h)-sqrt(x))/h-lim_(h->0)(2h)/h#

#color(white)(llllllll)=lim_(h->0)1/(h/(sqrt(x+h)-sqrt(x)))-2#

#color(white)(llllllll)=lim_(h->0)1/((h(sqrt(x+h)+sqrt(x)))/((sqrt(x+h)-sqrt(x))(sqrt(x+h)+sqrt(x))))-2#

#color(white)(llllllll)=lim_(h->0)((sqrt(x+h)-sqrt(x))(sqrt(x+h)+sqrt(x)))/(h(sqrt(x+h)+sqrt(x)))-2#

#color(white)(llllllll)=lim_(h->0)(x+h-x+sqrt(x^2+hx)-sqrt(x^2+hx))/(hsqrt(x+h)+hsqrt(x))-2#

#color(white)(llllllll)=lim_(h->0)h/(hsqrt(x+h)+hsqrt(x))-2#

#color(white)(llllllll)=lim_(h->0)1/(sqrt(x+h)+sqrt(x))-2#

#color(white)(llllllll)=1/(2sqrt(x))-2#