What is the derivative of f(x)=(\sqrt[3]{x})/(x-6) ?

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

What is the first derivative of f(x)=(\sqrt[3]{x})/(x-6) ?

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

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derivative of

Solution steps

Apply the Quotient Rule:

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

\frac{d}{d x} (x - 6) = 1

Simplify

x \to 3 lim (f (x) (x - 1))

\frac{\partial}{\partial y} (x - y)

x \to 0 lim (\frac{( 1 1 ^{x} - 1 )}{x})

\int \frac{5}{cos ( x ) - 1} d x

\int y cos (\frac{x}{y}) d x

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