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

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

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

The first derivative of f(t)=(\sqrt[3]{t})/(t-6) is (t-6-3\sqrt[3]{t}t^{2/3})/(3t^{2/3)(t-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 t} (t - 6) = 1

Simplify

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2}})

\int_{2}^{5} e^{- x} d x

\frac{\partial}{\partial y} (3 y^{2} cos (4 x))

(4 + x) \frac{d y}{d x} + y = 4 + x

\frac{\partial}{\partial y} (- 6 x y)

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