What is the derivative of y= 2/(sqrt(x)) ?

The derivative of y= 2/(sqrt(x)) is -1/(x^{3/2)}

What is the first derivative of y= 2/(sqrt(x)) ?

The first derivative of y= 2/(sqrt(x)) is -1/(x^{3/2)}

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derivative of

y = \frac{2}{x}

- \frac{1}{x ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (\frac{2}{x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (\frac{1}{x})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Simplify

2 (- \frac{1}{2} x^{- \frac{1}{2} - 1}) : - \frac{1}{x ^{\frac{3}{2}}}

= - \frac{1}{x ^{\frac{3}{2}}}

y = \frac{13}{ln ( x )}

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x \to 0 lim (x^{3} + 2 x^{2} - 4 x + 4)

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3 sin (x), 8 cos (x)

What is the derivative of y= 2/(sqrt(x)) ?
The derivative of y= 2/(sqrt(x)) is -1/(x^{3/2)}
What is the first derivative of y= 2/(sqrt(x)) ?
The first derivative of y= 2/(sqrt(x)) is -1/(x^{3/2)}