# How do you find the differential #dy# of the function #y=sqrtx+1/sqrtx#?

Note that

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To find the differential ( dy ) of the function ( y = \sqrt{x} + \frac{1}{\sqrt{x}} ), differentiate the function with respect to ( x ) and then multiply by ( dx ).

[ y = \sqrt{x} + \frac{1}{\sqrt{x}} ]

[ \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right) ]

Using the power rule and the chain rule, differentiate each term:

[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} ]

[ dy = \left(\frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}\right) dx ]

This is the differential of the function ( y = \sqrt{x} + \frac{1}{\sqrt{x}} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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