What is the derivative of f(x)=ln(x^2-4x+1) ?

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

What is the first derivative of f(x)=ln(x^2-4x+1) ?

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

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derivative of

f (x) = ln (x^{2} - 4 x + 1)

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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