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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y^{^{''}} + 4 y^{^{'}} + 30 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int sin (3 x) cos^{2} (3 x) d x

\int_{- 1}^{4} (5 x^{4} - 8 x^{3} + 6) d x

f^{^{'}} (x) = e^{x}

\int - 6 x e^{x} d x

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