What is the derivative of-2x^3+8/(x^2-6x-1) ?

The derivative of-2x^3+8/(x^2-6x-1) is -6x^2-(16)/(x^3)-6

What is the first derivative of-2x^3+8/(x^2-6x-1) ?

The first derivative of-2x^3+8/(x^2-6x-1) is -6x^2-(16)/(x^3)-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (- 2 x^{3} + \frac{8}{x ^{2}} - 6 x - 1)

- 6 x^{2} - \frac{16}{x ^{3}} - 6

Solution steps

\frac{d}{d x} (- 2 x^{3} + \frac{8}{x ^{2}} - 6 x - 1)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (\frac{8}{x ^{2}}) - \frac{d}{d x} (6 x) - \frac{d}{d x} (1)

\frac{d}{d x} (2 x^{3}) = 6 x^{2}

\frac{d}{d x} (\frac{8}{x ^{2}}) = - \frac{16}{x ^{3}}

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

\frac{d}{d x} (1) = 0

= - 6 x^{2} - \frac{16}{x ^{3}} - 6 - 0

Simplify

= - 6 x^{2} - \frac{16}{x ^{3}} - 6

\int - 8 x^{3} cos (5 x^{4} + 1) d x

3 e^{6 x}

\frac{\partial}{\partial x} (16 x - 4 y - 6 y x)

\int_{- 1}^{2} (2 - 5 x + \frac{1}{2} x^{2}) d x

f (x) = (x + 9)^{\frac{1}{2}}

What is the derivative of-2x^3+8/(x^2-6x-1) ?
The derivative of-2x^3+8/(x^2-6x-1) is -6x^2-(16)/(x^3)-6
What is the first derivative of-2x^3+8/(x^2-6x-1) ?
The first derivative of-2x^3+8/(x^2-6x-1) is -6x^2-(16)/(x^3)-6