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

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

The first derivative of f(t)=(\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_{0}^{5} π (25 - x^{2}) d x

x \to 4 lim ((x + 3)^{2})

\frac{d ^{2}}{d x ^{2}} (e^{- \frac{x ^{2}}{2}})

\frac{d ^{3}}{d x ^{3}} (sin (x))

x \to 0 + lim (tan (4 x))

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