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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

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

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

\int e^{- x} sin^{2} (x) d x

\frac{d x}{d t} = \frac{4 x}{t}, x (1) = 1

\int \frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{4 x} d x

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

\frac{4 x}{x + 1}

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