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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

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

y = 6 + 5 x^{2} + 2 x^{3}

\int \frac{2 x}{x ^{4} + 16} d x

3 cos (2 x), 3 - 3 cos (2 x)

x^{2} \cdot y^{^{'}} + (2 - y) = 0

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)