What is the derivative of #(sqrtx-1)/sqrtx#?

2 Answers
Jan 22, 2016

Answer:

#1/(2sqrt(x^3))#

Explanation:

The quotient rule states that

#d/dx[f(x)/g(x)]=(g(x)*f'(x)-f(x)*g'(x))/[g(x)]^2#

So application hereof in this particular case yields

#d/dx((sqrtx-1)/sqrtx)=((sqrtx*1/2x^(-1/2))-((sqrtx-1)1/2x^(-1/2)))/x#

#=(1/2-1/2+1/(2sqrtx))/x#

#=1/(2xsqrtx)#

Jan 22, 2016

Answer:

#1/(2x^(3/2))#

Explanation:

Begin by simplifying the function.

#f(x)=sqrtx/sqrtx-1/sqrtx=1-x^(-1/2)#

To differentiate from here, simply use the power rule.

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

This can also be written as

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