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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

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

\int 3 x e^{7 x} d x

(y - (x^{2} + y^{2})) d x - x d y = 0

\frac{\partial}{\partial y} (ln (x + y + z))

\frac{5 e ^{- s}}{s + 1}

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