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}

y = x^{2} + 5 x + 6

f (x) = x x, at x = 9

\int cos^{(5)} (5 x) sin (5 x) d x

\int_{0}^{9} 3 x - x d x

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

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)