What is the derivative of x+ln|x^2-4x+3| ?

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

What is the first derivative of x+ln|x^2-4x+3| ?

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d x}{d x} = 1

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

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

\int \frac{1}{1 + x} d x

\frac{\partial}{\partial x} (x x^{2} + y^{2})

y = 3 x, y = \frac{3}{x}

\frac{\partial}{\partial z} (4 x^{\frac{y}{z}})

2 x - 10, [4, 8]

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