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

The derivative of f(x)=(x^3)/(3-x^2) is (9x^2-x^4)/((3-x^2)^2)

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

The first derivative of f(x)=(x^3)/(3-x^2) is (9x^2-x^4)/((3-x^2)^2)

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derivative of

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

\frac{9 x ^{2} - x ^{4}}{( 3 - x ^{2} ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{3 x ^{2} ( 3 - x ^{2} ) - ( - 2 x ) x ^{3}}{( 3 - x ^{2} ) ^{2}} : \frac{9 x ^{2} - x ^{4}}{( 3 - x ^{2} ) ^{2}}

= \frac{9 x ^{2} - x ^{4}}{( 3 - x ^{2} ) ^{2}}

\int \frac{( 4 x - 5 ) ( 2 x - 7 )}{3} d x

\frac{\partial}{\partial x} (ln (x^{2} + 6 y^{2} + 8))

\frac{d y}{d x} = (y + 8)

\int \frac{sec ( θ )}{tan ( θ )} d θ

\frac{d y}{d x} + x \cdot y = - x^{2}

What is the derivative of f(x)=(x^3)/(3-x^2) ?
The derivative of f(x)=(x^3)/(3-x^2) is (9x^2-x^4)/((3-x^2)^2)
What is the first derivative of f(x)=(x^3)/(3-x^2) ?
The first derivative of f(x)=(x^3)/(3-x^2) is (9x^2-x^4)/((3-x^2)^2)