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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

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

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

\frac{( 1 + 16 t ^{2} )}{( 4 t ) ^{3}}

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 3)

\int_{- 1}^{1} x^{2} x^{3} + 1 d x

\int \frac{x ^{2} + 3 x}{x + 4} d x

\int (\frac{e ^{4 x} - 5}{e ^{2 x}}) d x

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