# How do you find the derivative of #y = (x^2 + sqrtx + 1 ) / x^(3/2)#?

Simplify first:

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To find the derivative of ( y = \frac{x^2 + \sqrt{x} + 1}{x^{3/2}} ), you can use the quotient rule. The quotient rule states that if ( y = \frac{u}{v} ), where ( u ) and ( v ) are functions of ( x ), then the derivative of ( y ) with respect to ( x ) is given by:

[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]

First, identify ( u ) and ( v ) in the given function:

( u = x^2 + \sqrt{x} + 1 )

( v = x^{3/2} )

Now, find the derivatives of ( u ) and ( v ) with respect to ( x ):

[ \frac{du}{dx} = 2x + \frac{1}{2\sqrt{x}} ]

[ \frac{dv}{dx} = \frac{3}{2}x^{1/2} ]

Now, substitute these derivatives into the quotient rule formula:

[ \frac{dy}{dx} = \frac{x^{3/2} (2x + \frac{1}{2\sqrt{x}}) - (x^2 + \sqrt{x} + 1) \frac{3}{2}x^{1/2}}{(x^{3/2})^2} ]

[ \frac{dy}{dx} = \frac{2x^{5/2} + \frac{1}{2} - \frac{3x^{3/2}}{2} - \frac{3\sqrt{x}}{2} - \frac{3}{2}x^{1/2}}{x^3} ]

[ \frac{dy}{dx} = \frac{2x^{5/2} - \frac{3x^{3/2}}{2} - \frac{3\sqrt{x}}{2} + \frac{1}{2}}{x^3} ]

This expression is the derivative of ( y ) with respect to ( 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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