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

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

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

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

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

\frac{4 x - 3}{2 x ^{2} - 3 x + 1}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{4 x - 3}{2 x ^{2} - 3 x + 1}

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