Question

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

Answer 1

`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)`

Answer 2

`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)`