What is the derivative of (4x/(2x^2+1)) ?

The derivative of (4x/(2x^2+1)) is (4(-2x^2+1))/((2x^2+1)^2)

What is the first derivative of (4x/(2x^2+1)) ?

The first derivative of (4x/(2x^2+1)) is (4(-2x^2+1))/((2x^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{4 x}{2 x ^{2} + 1})

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

= 4 \cdot \frac{1 \cdot ( 2 x ^{2} + 1 ) - 4 xx}{( 2 x ^{2} + 1 ) ^{2}}

Simplify

4 \cdot \frac{1 \cdot ( 2 x ^{2} + 1 ) - 4 xx}{( 2 x ^{2} + 1 ) ^{2}} : \frac{4 ( - 2 x ^{2} + 1 )}{( 2 x ^{2} + 1 ) ^{2}}

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

d er i v a t i v e (x + 2)^{2} + (10 - 3 x)^{2}

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

\int x^{2} - 6 x - 27 d x

4 x \frac{d y}{d x} = x^{2} - 9

\int_{- 1}^{1} ln (ln (x + 3)) d x

What is the derivative of (4x/(2x^2+1)) ?
The derivative of (4x/(2x^2+1)) is (4(-2x^2+1))/((2x^2+1)^2)
What is the first derivative of (4x/(2x^2+1)) ?
The first derivative of (4x/(2x^2+1)) is (4(-2x^2+1))/((2x^2+1)^2)