How do you use the limit definition of the derivative to find the derivative of #f(x)=1/sqrtx#?

1 Answer
Apr 25, 2017

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

Explanation:

By definition of derivative we have:

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

So for #f(x) =1/sqrtx#:

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

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

Rationalize the numerator using the identity: #(a+b)(a-b) = a^2-b^2#:

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

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

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

#f'(x) = -1/((sqrtxsqrt(x))(sqrtx+sqrt(x)))#

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