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

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

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

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

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

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

Solution steps

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

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

Simplify

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

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

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

sin (x + \frac{5}{2})

\frac{d y}{d x} = \frac{4}{x ^{2} - 1}

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

\frac{d}{d x} (6 e)

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