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

The derivative of (\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 (\sqrt[3]{x}/(x-6)) ?

The first derivative of (\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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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

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

ln (1 - x), 0

\int x (3 x + 1)^{7} d x

(t + 5)^{3}

\frac{d}{d x} (\frac{ln ( x )}{x ^{13}})

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