What is the derivative of 1/(2y^{3/2)} ?

The derivative of 1/(2y^{3/2)} is -3/(4y^{5/2)}

What is the first derivative of 1/(2y^{3/2)} ?

The first derivative of 1/(2y^{3/2)} is -3/(4y^{5/2)}

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derivative

- \frac{3}{4 y ^{\frac{5}{2}}}

Solution steps

\frac{d}{d y} (\frac{1}{2 y ^{\frac{3}{2}}})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d y} (\frac{1}{y ^{\frac{3}{2}}})

Apply exponent rule:

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

= \frac{1}{2} \frac{d}{d y} (y^{- \frac{3}{2}})

Apply the Power Rule:

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

= \frac{1}{2} (- \frac{3}{2} y^{- \frac{3}{2} - 1})

Simplify

\frac{1}{2} (- \frac{3}{2} y^{- \frac{3}{2} - 1}) : - \frac{3}{4 y ^{\frac{5}{2}}}

= - \frac{3}{4 y ^{\frac{5}{2}}}

What is the derivative of 1/(2y^{3/2)} ?
The derivative of 1/(2y^{3/2)} is -3/(4y^{5/2)}
What is the first derivative of 1/(2y^{3/2)} ?
The first derivative of 1/(2y^{3/2)} is -3/(4y^{5/2)}