What is the derivative of \sqrt[3]{x}-3/x ?

The derivative of \sqrt[3]{x}-3/x is (x^2+9x^{2/3})/(3x^{8/3)}

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

The first derivative of \sqrt[3]{x}-3/x is (x^2+9x^{2/3})/(3x^{8/3)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{x ^{2} + 9 x ^{\frac{2}{3}}}{3 x ^{\frac{8}{3}}}

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d x} (\frac{3}{x}) = - \frac{3}{x ^{2}}

= \frac{1}{3 x ^{\frac{2}{3}}} - (- \frac{3}{x ^{2}})

Simplify

\frac{1}{3 x ^{\frac{2}{3}}} - (- \frac{3}{x ^{2}}) : \frac{x ^{2} + 9 x ^{\frac{2}{3}}}{3 x ^{\frac{8}{3}}}

= \frac{x ^{2} + 9 x ^{\frac{2}{3}}}{3 x ^{\frac{8}{3}}}

(- 4 - 4) (- 42)

\int 5 x^{2} e^{x} d x

(x + \frac{2 y ^{2}}{x}) d x = y d y

(\frac{e ^{x}}{x})^{^{'}}

f (t) = (3 t + 14)^{\frac{2}{3}}

What is the derivative of \sqrt[3]{x}-3/x ?
The derivative of \sqrt[3]{x}-3/x is (x^2+9x^{2/3})/(3x^{8/3)}
What is the first derivative of \sqrt[3]{x}-3/x ?
The first derivative of \sqrt[3]{x}-3/x is (x^2+9x^{2/3})/(3x^{8/3)}