What is the derivative of y=(x^4)/(1-x^3) ?

The derivative of y=(x^4)/(1-x^3) is (4x^3-x^6)/((1-x^3)^2)

What is the first derivative of y=(x^4)/(1-x^3) ?

The first derivative of y=(x^4)/(1-x^3) is (4x^3-x^6)/((1-x^3)^2)

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derivative of

y = \frac{x ^{4}}{1 - x ^{3}}

\frac{4 x ^{3} - x ^{6}}{( 1 - x ^{3} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{4}}{1 - 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} ) ( 1 - x ^{3} ) - \frac{d}{d x} ( 1 - x ^{3} ) x ^{4}}{( 1 - x ^{3} ) ^{2}}

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

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

= \frac{4 x ^{3} ( 1 - x ^{3} ) - ( - 3 x ^{2} ) x ^{4}}{( 1 - x ^{3} ) ^{2}}

Simplify

\frac{4 x ^{3} ( 1 - x ^{3} ) - ( - 3 x ^{2} ) x ^{4}}{( 1 - x ^{3} ) ^{2}} : \frac{4 x ^{3} - x ^{6}}{( 1 - x ^{3} ) ^{2}}

= \frac{4 x ^{3} - x ^{6}}{( 1 - x ^{3} ) ^{2}}

x \to 5 + lim (\frac{x - 6}{x - 5})

\int 5 x^{- 6} d x

\frac{1}{8}

x \to 5 lim (\frac{9 x}{x ^{2} - 25})

\frac{\partial}{\partial y} (x y ln (y))

What is the derivative of y=(x^4)/(1-x^3) ?
The derivative of y=(x^4)/(1-x^3) is (4x^3-x^6)/((1-x^3)^2)
What is the first derivative of y=(x^4)/(1-x^3) ?
The first derivative of y=(x^4)/(1-x^3) is (4x^3-x^6)/((1-x^3)^2)