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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

3 y^{^{''}} + 8 y^{^{'}} - 16 y = 0

y^{^{''}} + 4 y = e^{- x}

\frac{\partial}{\partial x} (t^{2} e^{- x})

\int 2 x (x^{2} + 1)^{5} d x

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