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

The derivative of (x^3/(4-x^2)) is (12x^2-x^4)/((4-x^2)^2)

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

The first derivative of (x^3/(4-x^2)) is (12x^2-x^4)/((4-x^2)^2)

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

\frac{12 x ^{2} - x ^{4}}{( 4 - x ^{2} ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{3 x ^{2} ( 4 - x ^{2} ) - ( - 2 x ) x ^{3}}{( 4 - x ^{2} ) ^{2}} : \frac{12 x ^{2} - x ^{4}}{( 4 - x ^{2} ) ^{2}}

= \frac{12 x ^{2} - x ^{4}}{( 4 - x ^{2} ) ^{2}}

\frac{d}{d x} (sec (x) tan (x) x (x + 1))

x^{3} \frac{d y}{d x} + 4 x^{2} y = e^{- x}

\int 5 sin (3 x) sin (6 x) d x

\frac{15}{3 s + 3}

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

What is the derivative of (x^3/(4-x^2)) ?
The derivative of (x^3/(4-x^2)) is (12x^2-x^4)/((4-x^2)^2)
What is the first derivative of (x^3/(4-x^2)) ?
The first derivative of (x^3/(4-x^2)) is (12x^2-x^4)/((4-x^2)^2)