What is the derivative of (sqrt(x)/(\sqrt[3]{2x)}) ?

The derivative of (sqrt(x)/(\sqrt[3]{2x)}) is 1/(6\sqrt[3]{2)\sqrt[6]{x}x^{2/3}}

What is the first derivative of (sqrt(x)/(\sqrt[3]{2x)}) ?

The first derivative of (sqrt(x)/(\sqrt[3]{2x)}) is 1/(6\sqrt[3]{2)\sqrt[6]{x}x^{2/3}}

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

\frac{1}{6 3 2 6 x x ^{\frac{2}{3}}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{1}{6 3 2 6 x x ^{\frac{2}{3}}}

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 6} e^{4 t}

\frac{\partial}{\partial y} (8 x^{5 y})

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

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

\frac{d y}{d x} = x + y + 1

What is the derivative of (sqrt(x)/(\sqrt[3]{2x)}) ?
The derivative of (sqrt(x)/(\sqrt[3]{2x)}) is 1/(6\sqrt[3]{2)\sqrt[6]{x}x^{2/3}}
What is the first derivative of (sqrt(x)/(\sqrt[3]{2x)}) ?
The first derivative of (sqrt(x)/(\sqrt[3]{2x)}) is 1/(6\sqrt[3]{2)\sqrt[6]{x}x^{2/3}}