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

The derivative of 6/(\sqrt[3]{x}) is -2/(x^{4/3)}

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

The first derivative of 6/(\sqrt[3]{x}) is -2/(x^{4/3)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- \frac{2}{x ^{\frac{4}{3}}}

Solution steps

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 6 \frac{d}{d x} (x^{- \frac{1}{3}})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 6 (- \frac{1}{3} x^{- \frac{1}{3} - 1})

Simplify

6 (- \frac{1}{3} x^{- \frac{1}{3} - 1}) : - \frac{2}{x ^{\frac{4}{3}}}

= - \frac{2}{x ^{\frac{4}{3}}}

3 ln (5 x)

x^{2} + 2 x y + x \frac{d y}{d x} = 0

3 x^{4} - 65 x^{3}

\int 2 sec^{4} (x) d x

\int - \frac{5000}{x ^{2}} d x

What is the derivative of 6/(\sqrt[3]{x}) ?
The derivative of 6/(\sqrt[3]{x}) is -2/(x^{4/3)}
What is the first derivative of 6/(\sqrt[3]{x}) ?
The first derivative of 6/(\sqrt[3]{x}) is -2/(x^{4/3)}