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

The derivative of (\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 (\sqrt[3]{t})/(t-6) ?

The first derivative of (\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

(1 - x)^{\frac{1}{4}}

n = 1 \sum \infty (\frac{1}{8})^{n - 1}

\frac{d}{d x} (\frac{c e ^{x}}{1 + c e ^{x}})

y = x^{2} + 7, (3, 16)

\frac{d}{d x} (x^{3} + 2 x + 1)

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