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

The derivative of (x^4)/(2-x^3) is (8x^3-x^6)/((2-x^3)^2)

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

The first derivative of (x^4)/(2-x^3) is (8x^3-x^6)/((2-x^3)^2)

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derivative of

\frac{x ^{4}}{2 - x ^{3}}

\frac{8 x ^{3} - x ^{6}}{( 2 - x ^{3} ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{4 x ^{3} ( 2 - x ^{3} ) - ( - 3 x ^{2} ) x ^{4}}{( 2 - x ^{3} ) ^{2}} : \frac{8 x ^{3} - x ^{6}}{( 2 - x ^{3} ) ^{2}}

= \frac{8 x ^{3} - x ^{6}}{( 2 - x ^{3} ) ^{2}}

\frac{d}{d x} (47 x^{3} - 75 x)

y^{^{''}} + 7 y^{^{'}} + 10 y = 10 t e^{4 t}

\int_{0}^{1} \frac{57}{x ^{5}} d x

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

n = 0 \sum \infty (- 4)^{n}

What is the derivative of (x^4)/(2-x^3) ?
The derivative of (x^4)/(2-x^3) is (8x^3-x^6)/((2-x^3)^2)
What is the first derivative of (x^4)/(2-x^3) ?
The first derivative of (x^4)/(2-x^3) is (8x^3-x^6)/((2-x^3)^2)