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

The derivative of f(x)=(-3x^3-1)/(2x^2+1) is (-6x^4-9x^2+4x)/((2x^2+1)^2)

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

The first derivative of f(x)=(-3x^3-1)/(2x^2+1) is (-6x^4-9x^2+4x)/((2x^2+1)^2)

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derivative of

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

\frac{- 6 x ^{4} - 9 x ^{2} + 4 x}{( 2 x ^{2} + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (- 3 x^{3} - 1) = - 9 x^{2}

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

= \frac{( - 9 x ^{2} ) ( 2 x ^{2} + 1 ) - 4 x ( - 3 x ^{3} - 1 )}{( 2 x ^{2} + 1 ) ^{2}}

Simplify

\frac{( - 9 x ^{2} ) ( 2 x ^{2} + 1 ) - 4 x ( - 3 x ^{3} - 1 )}{( 2 x ^{2} + 1 ) ^{2}} : \frac{- 6 x ^{4} - 9 x ^{2} + 4 x}{( 2 x ^{2} + 1 ) ^{2}}

= \frac{- 6 x ^{4} - 9 x ^{2} + 4 x}{( 2 x ^{2} + 1 ) ^{2}}

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

\int \frac{5 x ^{3} - 4 x + 2}{x ^{3} - x ^{2}} d x

x \frac{d y}{d x} + y = 2 x + 1, y (1) = 6

x \to 1 lim (3 \cdot a + x)

\frac{\partial}{\partial y} (10 x z sin (y))

What is the derivative of f(x)=(-3x^3-1)/(2x^2+1) ?
The derivative of f(x)=(-3x^3-1)/(2x^2+1) is (-6x^4-9x^2+4x)/((2x^2+1)^2)
What is the first derivative of f(x)=(-3x^3-1)/(2x^2+1) ?
The first derivative of f(x)=(-3x^3-1)/(2x^2+1) is (-6x^4-9x^2+4x)/((2x^2+1)^2)