What is the derivative of f(x)=((x^3+1))/((x^4+1)) ?

The derivative of f(x)=((x^3+1))/((x^4+1)) is (-x^6-4x^3+3x^2)/((x^4+1)^2)

What is the first derivative of f(x)=((x^3+1))/((x^4+1)) ?

The first derivative of f(x)=((x^3+1))/((x^4+1)) is (-x^6-4x^3+3x^2)/((x^4+1)^2)

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{( x ^{3} + 1 )}{( x ^{4} + 1 )}

\frac{- x ^{6} - 4 x ^{3} + 3 x ^{2}}{( x ^{4} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 1}{x ^{4} + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} + 1 ) ( x ^{4} + 1 ) - \frac{d}{d x} ( x ^{4} + 1 ) ( x ^{3} + 1 )}{( x ^{4} + 1 ) ^{2}}

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

\frac{d}{d x} (x^{4} + 1) = 4 x^{3}

= \frac{3 x ^{2} ( x ^{4} + 1 ) - 4 x ^{3} ( x ^{3} + 1 )}{( x ^{4} + 1 ) ^{2}}

Expand

3 x^{2} (x^{4} + 1) - 4 x^{3} (x^{3} + 1) : - x^{6} - 4 x^{3} + 3 x^{2}

= \frac{- x ^{6} - 4 x ^{3} + 3 x ^{2}}{( x ^{4} + 1 ) ^{2}}

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

\int 2 4 - x^{2} d x

x \to π lim (\frac{( e ^{s i n (x)} - 1 )}{x - π})

x \to - \infty lim (7 x e^{x})

\frac{d y}{d x}, arctan (x y) = arcsin (2 x + 2 y), (0, 0)

What is the derivative of f(x)=((x^3+1))/((x^4+1)) ?
The derivative of f(x)=((x^3+1))/((x^4+1)) is (-x^6-4x^3+3x^2)/((x^4+1)^2)
What is the first derivative of f(x)=((x^3+1))/((x^4+1)) ?
The first derivative of f(x)=((x^3+1))/((x^4+1)) is (-x^6-4x^3+3x^2)/((x^4+1)^2)