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

The derivative of f(x)=(1+x)/((2x^2+1)^2) is (-6x^2-8x+1)/((2x^2+1)^3)

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

The first derivative of f(x)=(1+x)/((2x^2+1)^2) is (-6x^2-8x+1)/((2x^2+1)^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{- 6 x ^{2} - 8 x + 1}{( 2 x ^{2} + 1 ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} ((2 x^{2} + 1)^{2}) = 8 x (2 x^{2} + 1)

= \frac{1 \cdot ( 2 x ^{2} + 1 ) ^{2} - 8 x ( 2 x ^{2} + 1 ) ( 1 + x )}{( ( 2 x ^{2} + 1 ) ^{2} ) ^{2}}

Simplify

\frac{1 \cdot ( 2 x ^{2} + 1 ) ^{2} - 8 x ( 2 x ^{2} + 1 ) ( 1 + x )}{( ( 2 x ^{2} + 1 ) ^{2} ) ^{2}} : \frac{- 6 x ^{2} - 8 x + 1}{( 2 x ^{2} + 1 ) ^{3}}

= \frac{- 6 x ^{2} - 8 x + 1}{( 2 x ^{2} + 1 ) ^{3}}

\frac{d}{d t} (t^{3} - 5 t^{2} + 9 t)

y = 2 x + 2, y = x^{2} - 2

y^{2} = 2 x, y = - 4 + x

\int \frac{1}{25 - 4 x ^{2}} d x

\int t \cdot cos (t) d t

What is the derivative of f(x)=(1+x)/((2x^2+1)^2) ?
The derivative of f(x)=(1+x)/((2x^2+1)^2) is (-6x^2-8x+1)/((2x^2+1)^3)
What is the first derivative of f(x)=(1+x)/((2x^2+1)^2) ?
The first derivative of f(x)=(1+x)/((2x^2+1)^2) is (-6x^2-8x+1)/((2x^2+1)^3)