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

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

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

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

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derivative of

f (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}}

\int 1 + u^{2} d u

\int_{0}^{4} x^{3} x^{2} + 16 d x

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

t^{2} y^{^{'}} + t y = 5

x sin (3 x)

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