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

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Answer 1 · 2016-01-22T21:25:03

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

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 · 2016-01-22T22:04:27

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

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

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

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