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

The derivative of (\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 (\sqrt[3]{t})/(t-9) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{3}{8 + 6 x} d x

y = \frac{1}{x}, y = \frac{1}{4} x

x \to 1 lim (\frac{x + 2}{x ^{3} + 8})

\int \frac{2}{x ln ^{3} ( x )} d x

\frac{\partial}{\partial x} (x^{4} ln (yz))

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