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

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

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

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

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

\frac{6 x - 4}{3 x ^{2} - 4 x}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{6 x - 4}{3 x ^{2} - 4 x}

x \to 2 lim (\frac{x ^{2}}{x ^{2} + 4})

\int cos (x) + \frac{1}{8} x d x

f^{^{''}} (x) = - 2 + 36 x - 12 x^{2}

- 4 sin (2 x)

f (x) = arccos (e^{- x})

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