What is the derivative of-1/3 (x^{-3}-x^6) ?

The derivative of-1/3 (x^{-3}-x^6) is (2x^9+1)/(x^4)

What is the first derivative of-1/3 (x^{-3}-x^6) ?

The first derivative of-1/3 (x^{-3}-x^6) is (2x^9+1)/(x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{2 x ^{9} + 1}{x ^{4}}

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (x^{- 3}) = - \frac{3}{x ^{4}}

\frac{d}{d x} (x^{6}) = 6 x^{5}

= - \frac{1}{3} (- \frac{3}{x ^{4}} - 6 x^{5})

Simplify

- \frac{1}{3} (- \frac{3}{x ^{4}} - 6 x^{5}) : \frac{2 x ^{9} + 1}{x ^{4}}

= \frac{2 x ^{9} + 1}{x ^{4}}

33 + 4 t

\frac{x ^{2}}{( 4 + x )}

\int \frac{1}{1 + cos ( θ )} d θ

\frac{\partial}{\partial x} (3 x^{2} cos^{2} (x - y))

\frac{\partial}{\partial x} (x cos (x y) - 2 y)

What is the derivative of-1/3 (x^{-3}-x^6) ?
The derivative of-1/3 (x^{-3}-x^6) is (2x^9+1)/(x^4)
What is the first derivative of-1/3 (x^{-3}-x^6) ?
The first derivative of-1/3 (x^{-3}-x^6) is (2x^9+1)/(x^4)