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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

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

n = 0 \sum \infty \frac{n + 1}{2 n - 3}

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

\frac{d x}{d t} = 7 (x^{2} + 1), x (\frac{π}{4}) = 1

6 t^{2} - 4 t + 9

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