What is the derivative of y= 1/(sqrt(x)) ?

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

What is the first derivative of y= 1/(sqrt(x)) ?

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

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derivative of

y = \frac{1}{x}

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

Solution steps

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

\int 7 x^{2} - 3 x + 8 + \frac{1}{x} + \frac{2}{x ^{2}} d x

\int_{0}^{3} 10 - x - y^{2} d x

f (x) = x^{2} e^{3 x}

x \to 2 + lim (2 x - 3)

\frac{d}{d x} (e^{- x} - x^{2})

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