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

The derivative of (ln(3x^2-6x)/(4x^3-2)) is

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

The first derivative of (ln(3x^2-6x)/(4x^3-2)) is

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\int (ln (x))^{7} d x

\frac{d}{d x} (4 sec (4 x))

\int e^{(x + 1)} d x

n = 1 \sum \infty cos (\frac{1}{n})

x \frac{d y}{d x} + 6 y = x^{3} - x

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