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How do you find the derivative of #y=sqrtx/x#?

Answer 1

William Abdalla

#dy/dx=-1/2x^(-3/2)#

Using the power rule:

#dy/dx = d/dxsqrt(x)/x#

#=d/dxx^(1/2)/x^1#

#=d/dxx^(1/2-1)#

#=d/dxx^(-1/2)#

#=-1/2x^(-1/2-1)#

#=-1/2x^(-3/2)#

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Answer 2

Christopher Agresta

To find the derivative of ( y = \frac{\sqrt{x}}{x} ), use the quotient rule. The quotient rule states that if ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ). In this case, let ( u = \sqrt{x} ) and ( v = x ). Then, ( u' = \frac{1}{2\sqrt{x}} ) and ( v' = 1 ). Now apply the quotient rule:

[ y' = \frac{\frac{1}{2\sqrt{x}} \cdot x - \sqrt{x} \cdot 1}{x^2} ]

[ y' = \frac{\frac{x}{2\sqrt{x}} - \sqrt{x}}{x^2} ]

[ y' = \frac{\frac{x - 2\sqrt{x}}{2\sqrt{x}}}{x^2} ]

[ y' = \frac{x - 2\sqrt{x}}{2\sqrt{x} \cdot x^2} ]

[ y' = \frac{x - 2\sqrt{x}}{2x^{3/2}} ]

Therefore, the derivative of ( y = \frac{\sqrt{x}}{x} ) is ( y' = \frac{x - 2\sqrt{x}}{2x^{3/2}} ).

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Answer from HIX Tutor

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.