# How do you differentiate #f(t)=sqrtt-1/sqrtt#?

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To differentiate ( f(t) = \frac{\sqrt{t} - 1}{\sqrt{t}} ), you can use the quotient rule. The quotient rule states that if you have a function ( u(t) ) divided by another function ( v(t) ), then the derivative is given by:

[ \frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} ]

Using this rule, let's differentiate ( f(t) ):

[ u(t) = \sqrt{t} - 1 ] [ v(t) = \sqrt{t} ]

Now find the derivatives of ( u(t) ) and ( v(t) ):

[ u'(t) = \frac{1}{2\sqrt{t}} ] [ v'(t) = \frac{1}{2\sqrt{t}} ]

Then, apply the quotient rule:

[ f'(t) = \frac{(1/2\sqrt{t})\sqrt{t} - (\sqrt{t} - 1)(1/2\sqrt{t})}{(\sqrt{t})^2} ]

[ f'(t) = \frac{\frac{1}{2} - \frac{\sqrt{t}}{2} - \frac{1}{2\sqrt{t}}}{t} ]

[ f'(t) = \frac{1 - \sqrt{t} - 1}{2t\sqrt{t}} ]

[ f'(t) = \frac{-\sqrt{t}}{2t\sqrt{t}} ]

[ f'(t) = \frac{-1}{2t} ]

So, the derivative of ( f(t) = \frac{\sqrt{t} - 1}{\sqrt{t}} ) with respect to ( t ) is ( f'(t) = \frac{-1}{2t} ).

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