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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 0 lim (\frac{1}{x} ln (1 + x))

\int_{0}^{π} t^{2} e^{- s t} d t

y = e^{x}, y = e^{- x}, x = 1

x e^{- 5 x}

\int \frac{x ^{3} + 3 x ^{2} + 6}{x ^{2} + 2} d x

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