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

The derivative of (x^4/(3-x^3)) is (12x^3-x^6)/((3-x^3)^2)

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

The first derivative of (x^4/(3-x^3)) is (12x^3-x^6)/((3-x^3)^2)

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

\frac{12 x ^{3} - x ^{6}}{( 3 - x ^{3} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (x^{4}) = 4 x^{3}

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

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

Simplify

\frac{4 x ^{3} ( 3 - x ^{3} ) - ( - 3 x ^{2} ) x ^{4}}{( 3 - x ^{3} ) ^{2}} : \frac{12 x ^{3} - x ^{6}}{( 3 - x ^{3} ) ^{2}}

= \frac{12 x ^{3} - x ^{6}}{( 3 - x ^{3} ) ^{2}}

\int_{0}^{\frac{π}{3}} 1 sec^{2} (x) d x

\int \frac{5}{x ln ( 4 x )} d x

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

\frac{d}{d x} (7 x - 8)

n = 0 \sum \infty (400) (0.1)^{n}

What is the derivative of (x^4/(3-x^3)) ?
The derivative of (x^4/(3-x^3)) is (12x^3-x^6)/((3-x^3)^2)
What is the first derivative of (x^4/(3-x^3)) ?
The first derivative of (x^4/(3-x^3)) is (12x^3-x^6)/((3-x^3)^2)