What is the derivative of-1/(4sqrt(x^3)) ?

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

What is the first derivative of-1/(4sqrt(x^3)) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (x^{3}) = 3 x^{2}

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

Simplify

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

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

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

\frac{\partial}{\partial x} (sin (\frac{3 x}{y}))

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

\int \frac{x + 4}{x ^{2} + 8 x} d x

x y^{^{'}} + y = 2 x

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