What is the (-1/(4x^{3/2)})^' ?

The (-1/(4x^{3/2)})^' is 3/(8x^{5/2)}

What is the first (-1/(4x^{3/2)})^' ?

The first (-1/(4x^{3/2)})^' is 3/(8x^{5/2)}

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(- \frac{1}{4 x ^{\frac{3}{2}}})^{^{'}}

\frac{3}{8 x ^{\frac{5}{2}}}

Solution steps

(- \frac{1}{4 x ^{\frac{3}{2}}})^{'}

Take the constant out:

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

= - \frac{1}{4} (\frac{1}{x ^{\frac{3}{2}}})^{^{'}}

Apply exponent rule:

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

= - \frac{1}{4} (x^{- \frac{3}{2}})^{^{'}}

Apply the Power Rule:

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

= - \frac{1}{4} (- \frac{3}{2} x^{- \frac{3}{2} - 1})

Simplify

- \frac{1}{4} (- \frac{3}{2} x^{- \frac{3}{2} - 1}) : \frac{3}{8 x ^{\frac{5}{2}}}

= \frac{3}{8 x ^{\frac{5}{2}}}

\frac{d y}{d x} + y^{9} x + 9 y = 0

x \to 4 lim (x^{2} - 2 x - 8)

\frac{d}{d x} ((6 x^{2} - 6 x + 3) (1 + 2 x))

(2.2) (4.3)

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

What is the (-1/(4x^{3/2)})^' ?
The (-1/(4x^{3/2)})^' is 3/(8x^{5/2)}
What is the first (-1/(4x^{3/2)})^' ?
The first (-1/(4x^{3/2)})^' is 3/(8x^{5/2)}