What is the derivative of ln((x^3-2x/(4+x))) ?

The derivative of ln((x^3-2x/(4+x))) is (2x^3+12x^2-8)/((x^3-2x)(x+4))

What is the first derivative of ln((x^3-2x/(4+x))) ?

The first derivative of ln((x^3-2x/(4+x))) is (2x^3+12x^2-8)/((x^3-2x)(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} (ln (\frac{x ^{3} - 2 x}{4 + x}))

\frac{2 x ^{3} + 12 x ^{2} - 8}{( x ^{3} - 2 x ) ( x + 4 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{3} - 2 x}{4 + x}) = \frac{2 x ^{3} + 12 x ^{2} - 8}{( 4 + x ) ^{2}}

= \frac{1}{\frac{x ^{3} - 2 x}{4 + x}} \cdot \frac{2 x ^{3} + 12 x ^{2} - 8}{( 4 + x ) ^{2}}

Simplify

\frac{1}{\frac{x ^{3} - 2 x}{4 + x}} \cdot \frac{2 x ^{3} + 12 x ^{2} - 8}{( 4 + x ) ^{2}} : \frac{2 x ^{3} + 12 x ^{2} - 8}{( x ^{3} - 2 x ) ( x + 4 )}

= \frac{2 x ^{3} + 12 x ^{2} - 8}{( x ^{3} - 2 x ) ( x + 4 )}

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

\frac{d}{d x} (\frac{x}{2})

11 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 9

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

- sin (\frac{x}{2})

What is the derivative of ln((x^3-2x/(4+x))) ?
The derivative of ln((x^3-2x/(4+x))) is (2x^3+12x^2-8)/((x^3-2x)(x+4))
What is the first derivative of ln((x^3-2x/(4+x))) ?
The first derivative of ln((x^3-2x/(4+x))) is (2x^3+12x^2-8)/((x^3-2x)(x+4))