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

The derivative of f(x)=(\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 f(x)=(\sqrt[3]{x})/(x-9) ?

The first derivative of f(x)=(\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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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 - 9) = 1

Simplify

\frac{\partial}{\partial x} (\frac{1}{∣ x ∣})

\int_{0}^{2} 2 x + 5 d x

\frac{d}{d x} (40 x (5 x^{2} - 8))

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

\int x^{- \frac{1}{3}} d x

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