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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 3 \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}}

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

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

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

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

Simplify

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

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

y = \frac{x - 2}{x + 2}

f (x) = 7 x e^{x}, at x = 0

\frac{d}{d x} (5 x^{\frac{3}{5}} - 4)

y = \frac{6 + x f ( x )}{x}

\int y^{2} 2 y^{3} + 1 d y

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