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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

2 \cdot \frac{1 \cdot ( 3 x ^{2} - 3 ) - 6 xx}{( 3 x ^{2} - 3 ) ^{2}} : - \frac{2 ( x ^{2} + 1 )}{3 ( x ^{2} - 1 ) ^{2}}

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

t an g e n t f (x) = 6 x, at x = 9

\int 2 e^{x} x d x

d er i v a t i v e y = ln (x^{5})

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

f (x) = \frac{x ^{2}}{3}

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