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

The derivative of (x^4)/(4-x^3) is (16x^3-x^6)/((4-x^3)^2)

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

The first derivative of (x^4)/(4-x^3) is (16x^3-x^6)/((4-x^3)^2)

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derivative of

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

\frac{16 x ^{3} - x ^{6}}{( 4 - x ^{3} ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{4 x ^{3} ( 4 - x ^{3} ) - ( - 3 x ^{2} ) x ^{4}}{( 4 - x ^{3} ) ^{2}} : \frac{16 x ^{3} - x ^{6}}{( 4 - x ^{3} ) ^{2}}

= \frac{16 x ^{3} - x ^{6}}{( 4 - x ^{3} ) ^{2}}

\frac{\partial}{\partial s} (t - 2 s)

\int \frac{x + 5}{( x + 4 ) ( x ^{2} - 1 )} d x

\int 25 - x^{2} (2 x) d x

\frac{\partial}{\partial x} (\frac{- t ^{2}}{( 1 - x ^{2} )})

y^{^{'}} - 2 y = x + 5

What is the derivative of (x^4)/(4-x^3) ?
The derivative of (x^4)/(4-x^3) is (16x^3-x^6)/((4-x^3)^2)
What is the first derivative of (x^4)/(4-x^3) ?
The first derivative of (x^4)/(4-x^3) is (16x^3-x^6)/((4-x^3)^2)