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

The derivative of (2x+3/(3x^2-4)) is (-6x^2-18x-8)/((3x^2-4)^2)

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

The first derivative of (2x+3/(3x^2-4)) is (-6x^2-18x-8)/((3x^2-4)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\frac{- 6 x ^{2} - 18 x - 8}{( 3 x ^{2} - 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (2 x + 3) = 2

\frac{d}{d x} (3 x^{2} - 4) = 6 x

= \frac{2 ( 3 x ^{2} - 4 ) - 6 x ( 2 x + 3 )}{( 3 x ^{2} - 4 ) ^{2}}

Expand

2 (3 x^{2} - 4) - 6 x (2 x + 3) : - 6 x^{2} - 18 x - 8

= \frac{- 6 x ^{2} - 18 x - 8}{( 3 x ^{2} - 4 ) ^{2}}

What is the derivative of (2x+3/(3x^2-4)) ?
The derivative of (2x+3/(3x^2-4)) is (-6x^2-18x-8)/((3x^2-4)^2)
What is the first derivative of (2x+3/(3x^2-4)) ?
The first derivative of (2x+3/(3x^2-4)) is (-6x^2-18x-8)/((3x^2-4)^2)