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

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

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

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

Simplify

f (x) = x^{3} + 3 x^{2} - 2 x + 1

x \to + 7 + lim (\frac{5}{x - 7})

\frac{1}{3}

\int_{0}^{8} (x^{2} + 8) e^{- x} d x

\int_{1}^{2} (x - 1) 2 - x d x

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