What is the derivative of (\sqrt[3]{x}/(x-9)) ?

The derivative of (\sqrt[3]{x}/(x-9)) is (x-9-3\sqrt[3]{x}x^{2/3})/(3x^{2/3)(x-9)^2}

What is the first derivative of (\sqrt[3]{x}/(x-9)) ?

The first derivative of (\sqrt[3]{x}/(x-9)) is (x-9-3\sqrt[3]{x}x^{2/3})/(3x^{2/3)(x-9)^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 - 9) = 1

Simplify

X^{3} - X^{2} + 4 X + 5

f (x) = \frac{125}{x ^{2} + 24}, at x = - 1

\frac{d}{d x} (cos (9 x^{2}))

\int \frac{1}{2 x ^{2} - 4 x + 20} d x

\frac{\partial}{\partial x} (x^{2} e^{y z^{2}})

What is the derivative of (\sqrt[3]{x}/(x-9)) ?
The derivative of (\sqrt[3]{x}/(x-9)) is (x-9-3\sqrt[3]{x}x^{2/3})/(3x^{2/3)(x-9)^2}
What is the first derivative of (\sqrt[3]{x}/(x-9)) ?
The first derivative of (\sqrt[3]{x}/(x-9)) is (x-9-3\sqrt[3]{x}x^{2/3})/(3x^{2/3)(x-9)^2}