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

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

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

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

Simplify

\int - \frac{1}{2} d x

\int_{0}^{1} x - x^{2} d x

\int (y - 1) (2 y + 1) d y

\frac{d y}{d x} + 2 x y = 2 x^{3}

\frac{3 ( s + 3 )}{s ^{2} + 6 s + 8}

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