What is the derivative of (-2sqrt(x)/(-4x^4)) ?

The derivative of (-2sqrt(x)/(-4x^4)) is -7/(4x^{9/2)}

What is the first derivative of (-2sqrt(x)/(-4x^4)) ?

The first derivative of (-2sqrt(x)/(-4x^4)) is -7/(4x^{9/2)}

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

- \frac{7}{4 x ^{\frac{9}{2}}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{- 2}{- 4} \cdot \frac{\frac{1}{2 x} x ^{4} - 4 x ^{3} x}{( x ^{4} ) ^{2}} : - \frac{7}{4 x ^{\frac{9}{2}}}

= - \frac{7}{4 x ^{\frac{9}{2}}}

\int_{0}^{2} \frac{1}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (x^{4} y^{3} + 8 x^{2} y)

\frac{d}{d x} (\frac{x - 5}{5})

81 - y^{2}

y^{^{'}} = 6 - 3 y

What is the derivative of (-2sqrt(x)/(-4x^4)) ?
The derivative of (-2sqrt(x)/(-4x^4)) is -7/(4x^{9/2)}
What is the first derivative of (-2sqrt(x)/(-4x^4)) ?
The first derivative of (-2sqrt(x)/(-4x^4)) is -7/(4x^{9/2)}