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

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

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

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

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derivative of

y = \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}}

x d x = y d y

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

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

\int_{\frac{1}{5}}^{7} 16 x ln (5 x) d x

\int_{2}^{4} (x^{2} + 4)^{3} x d x

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