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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(2 - 2 x) (3 x^{2} + 1) - 6 x (2 x - x^{2}) : - 6 x^{2} - 2 x + 2

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

(- 5) (0.1)

x y^{2} y^{^{'}} = y^{3} - x^{3}, y (1) = 3

x \to 32 lim (\frac{x ^{2}}{4} - \frac{8}{x})

((- 3 cos (2 x) - sin (3 x) - 4)^{2})^{^{'}}

\int 6 - x d x

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