What is the first (2/(x^3))^' ?

The first (2/(x^3))^' is -6/(x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2/(x^3))^'

(\frac{2}{x ^{3}})^{^{'}}

- \frac{6}{x ^{4}}

Solution steps

(\frac{2}{x ^{3}})^{'}

Take the constant out:

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

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

Apply exponent rule:

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

= 2 (x^{- 3})^{^{'}}

Apply the Power Rule:

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

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

Simplify

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

= - \frac{6}{x ^{4}}

y = \frac{1}{3} x^{3}

\int e^{- 0.2 x} d x

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{15}{2}}} d x

\int 5 (x^{5} - 10 x)^{2} (x^{4} - 2) d x

\frac{d}{d x} ((4 x^{2} - 1) (7 x^{3} + x))