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

The derivative of (6x-1/(3x^2-x)) is (-18x^2+6x-1)/((3x^2-x)^2)

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

The first derivative of (6x-1/(3x^2-x)) is (-18x^2+6x-1)/((3x^2-x)^2)

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

\frac{- 18 x ^{2} + 6 x - 1}{( 3 x ^{2} - x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{6 ( 3 x ^{2} - x ) - ( 6 x - 1 ) ( 6 x - 1 )}{( 3 x ^{2} - x ) ^{2}} : \frac{- 18 x ^{2} + 6 x - 1}{( 3 x ^{2} - x ) ^{2}}

= \frac{- 18 x ^{2} + 6 x - 1}{( 3 x ^{2} - x ) ^{2}}

\int \frac{5 x ^{4}}{x ^{5} + 8} d x

\int_{1}^{3} \frac{x}{x ^{2} + 1} d x

\frac{d}{d x} (lo g_{8} (x + 9))

\int 13 sin^{2} (x co) s^{2} x d x

\int \frac{1}{4 e ^{- 9 x} + e ^{9 x}} d x

What is the derivative of (6x-1/(3x^2-x)) ?
The derivative of (6x-1/(3x^2-x)) is (-18x^2+6x-1)/((3x^2-x)^2)
What is the first derivative of (6x-1/(3x^2-x)) ?
The first derivative of (6x-1/(3x^2-x)) is (-18x^2+6x-1)/((3x^2-x)^2)