What is the derivative of (2x+x^4/((1-x^3)^2)) ?

The derivative of (2x+x^4/((1-x^3)^2)) is (2(x^6+7x^3+1))/((1-x^3)^3)

What is the first derivative of (2x+x^4/((1-x^3)^2)) ?

The first derivative of (2x+x^4/((1-x^3)^2)) is (2(x^6+7x^3+1))/((1-x^3)^3)

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

\frac{2 ( x ^{6} + 7 x ^{3} + 1 )}{( 1 - x ^{3} ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 2 + 4 x ^{3} ) ( 1 - x ^{3} ) ^{2} - ( - 6 x ^{2} ( 1 - x ^{3} ) ) ( 2 x + x ^{4} )}{( ( 1 - x ^{3} ) ^{2} ) ^{2}} : \frac{2 ( x ^{6} + 7 x ^{3} + 1 )}{( 1 - x ^{3} ) ^{3}}

= \frac{2 ( x ^{6} + 7 x ^{3} + 1 )}{( 1 - x ^{3} ) ^{3}}

\int ln (sin (x)) d x

\frac{\partial}{\partial y} (2 e^{- l n (y)} y)

\frac{d ^{2}}{d x ^{2}} (cos (- 1.5 x))

\int 25 - t^{2} d t

\int x^{8} e^{- x^{9}} d x

What is the derivative of (2x+x^4/((1-x^3)^2)) ?
The derivative of (2x+x^4/((1-x^3)^2)) is (2(x^6+7x^3+1))/((1-x^3)^3)
What is the first derivative of (2x+x^4/((1-x^3)^2)) ?
The first derivative of (2x+x^4/((1-x^3)^2)) is (2(x^6+7x^3+1))/((1-x^3)^3)