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

The derivative of (3x-4/(6x^2-1)) is (-18x^2+48x-3)/((6x^2-1)^2)

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

The first derivative of (3x-4/(6x^2-1)) is (-18x^2+48x-3)/((6x^2-1)^2)

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

\frac{- 18 x ^{2} + 48 x - 3}{( 6 x ^{2} - 1 ) ^{2}}

Solution steps

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

\frac{d}{d x} (3 x - 4) = 3

\frac{d}{d x} (6 x^{2} - 1) = 12 x

= \frac{3 ( 6 x ^{2} - 1 ) - 12 x ( 3 x - 4 )}{( 6 x ^{2} - 1 ) ^{2}}

Expand

3 (6 x^{2} - 1) - 12 x (3 x - 4) : - 18 x^{2} + 48 x - 3

= \frac{- 18 x ^{2} + 48 x - 3}{( 6 x ^{2} - 1 ) ^{2}}

f (x) = ln (5 x - 7)

\int \frac{x ^{2} + 9}{x ^{2} + 4} d x

\int \frac{1}{7 y} d y

\frac{d y}{d x} = 1 - \frac{y}{x}

\frac{d}{d x} ((arctanh (3 x^{2}))^{3})

What is the derivative of (3x-4/(6x^2-1)) ?
The derivative of (3x-4/(6x^2-1)) is (-18x^2+48x-3)/((6x^2-1)^2)
What is the first derivative of (3x-4/(6x^2-1)) ?
The first derivative of (3x-4/(6x^2-1)) is (-18x^2+48x-3)/((6x^2-1)^2)