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

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

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

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

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{x}{1 + x ^{2} + x ^{4}})

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (- 7 x^{3} - \frac{2}{x ^{3}} + 7 x)

\int \frac{1}{25 x ^{2} - 81} d x

y^{^{'}} - y^{2} = 0

sin^{3} (cos (8 x))

\int cos^{2} (2 x) cos (2 x) d x

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