How do you find the derivative of $y = \frac{\sqrt{x}}{x}$ $y=sqrtx/x$ ?

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How do you find the derivative of $y = \frac{\sqrt{x}}{x}$ $y=sqrtx/x$ ?

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Answer 1 · 2017-01-03T02:19:37

$\frac{d y}{d x} = - \frac{1}{2} x^{- \frac{3}{2}}$ $dy/dx=-1/2x^(-3/2)$

Explanation:

Using the power rule:

$\frac{d y}{d x} = \frac{d}{d x} \frac{\sqrt{x}}{x}$ $dy/dx = d/dxsqrt(x)/x$

$= \frac{d}{d x} \frac{x^{\frac{1}{2}}}{x^{1}}$ $=d/dxx^(1/2)/x^1$

$= \frac{d}{d x} x^{\frac{1}{2} - 1}$ $=d/dxx^(1/2-1)$

$= \frac{d}{d x} x^{- \frac{1}{2}}$ $=d/dxx^(-1/2)$

$= - \frac{1}{2} x^{- \frac{1}{2} - 1}$ $=-1/2x^(-1/2-1)$

$= - \frac{1}{2} x^{- \frac{3}{2}}$ $=-1/2x^(-3/2)$

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How do you find the derivative of $y = \frac{\sqrt{x}}{x}$ $y=sqrtx/x$ ?

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Explanation:

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