How do I find derivatives of radicals like #sqrt(x)#?

1 Answer
Oct 11, 2014

We can use the Power Rule and the Difference Quotient ( First Principles ).

Power Rule

#f(x)=sqrt(x)=x^(1/2)#

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

Difference Quotient ( First Principles )

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

#f(x)=sqrt(x)#

#f(x+h)=sqrt(x+h)#

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

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

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

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

#f'(x)=lim_(h->0)1/(sqrt(x+h)+sqrt(x))#

#f'(x)=1/(sqrt(x+0)+sqrt(x))#

#f'(x)=1/(sqrt(x)+sqrt(x))#

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

Please see the videos below and a similar question.

Difference Quotient

Power Rule

How do you find the derivative of #f(x)=sqrt(x)# ?