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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}}

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

x = 1, x = 6 - y^{2}

\int_{a}^{b} 5 x^{3} d x

\frac{\partial}{\partial x} (x x)

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