What is the derivative of ((x^3-6x)^9)/9 ?

The derivative of ((x^3-6x)^9)/9 is (x^3-6x)^8(3x^2-6)

What is the first derivative of ((x^3-6x)^9)/9 ?

The first derivative of ((x^3-6x)^9)/9 is (x^3-6x)^8(3x^2-6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{( x ^{3} - 6 x ) ^{9}}{9}

(x^{3} - 6 x)^{8} (3 x^{2} - 6)

Solution steps

\frac{d}{d x} (\frac{( x ^{3} - 6 x ) ^{9}}{9})

Take the constant out:

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

= \frac{1}{9} \frac{d}{d x} ((x^{3} - 6 x)^{9})

Apply the chain rule:

9 (x^{3} - 6 x)^{8} \frac{d}{d x} (x^{3} - 6 x)

= 9 (x^{3} - 6 x)^{8} \frac{d}{d x} (x^{3} - 6 x)

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

= \frac{1}{9} \cdot 9 (x^{3} - 6 x)^{8} (3 x^{2} - 6)

Simplify

\frac{1}{9} \cdot 9 (x^{3} - 6 x)^{8} (3 x^{2} - 6) : (x^{3} - 6 x)^{8} (3 x^{2} - 6)

= (x^{3} - 6 x)^{8} (3 x^{2} - 6)

\int \frac{( x ^{3} )}{1 + x ^{2}} d x

\frac{\partial}{\partial x} (2 x^{3} y + 4)

x \to 3 + lim (\frac{1}{x - 3})

\int_{- \infty}^{0} \frac{1}{5 - 3 x} d x

y^{^{'}} + y tan (t) = \frac{tan ( t )}{cos ( t )}

What is the derivative of ((x^3-6x)^9)/9 ?
The derivative of ((x^3-6x)^9)/9 is (x^3-6x)^8(3x^2-6)
What is the first derivative of ((x^3-6x)^9)/9 ?
The first derivative of ((x^3-6x)^9)/9 is (x^3-6x)^8(3x^2-6)