What is the derivative of f(x)=ln(x^3-5x^2+3x-1) ?

The derivative of f(x)=ln(x^3-5x^2+3x-1) is (3x^2-10x+3)/(x^3-5x^2+3x-1)

What is the first derivative of f(x)=ln(x^3-5x^2+3x-1) ?

The first derivative of f(x)=ln(x^3-5x^2+3x-1) is (3x^2-10x+3)/(x^3-5x^2+3x-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^{3} - 5 x^{2} + 3 x - 1)

\frac{3 x ^{2} - 10 x + 3}{x ^{3} - 5 x ^{2} + 3 x - 1}

Solution steps

\frac{d}{d x} (ln (x^{3} - 5 x^{2} + 3 x - 1))

Apply the chain rule:

\frac{1}{x ^{3} - 5 x ^{2} + 3 x - 1} \frac{d}{d x} (x^{3} - 5 x^{2} + 3 x - 1)

= \frac{1}{x ^{3} - 5 x ^{2} + 3 x - 1} \frac{d}{d x} (x^{3} - 5 x^{2} + 3 x - 1)

\frac{d}{d x} (x^{3} - 5 x^{2} + 3 x - 1) = 3 x^{2} - 10 x + 3

= \frac{1}{x ^{3} - 5 x ^{2} + 3 x - 1} (3 x^{2} - 10 x + 3)

Simplify

\frac{1}{x ^{3} - 5 x ^{2} + 3 x - 1} (3 x^{2} - 10 x + 3) : \frac{3 x ^{2} - 10 x + 3}{x ^{3} - 5 x ^{2} + 3 x - 1}

= \frac{3 x ^{2} - 10 x + 3}{x ^{3} - 5 x ^{2} + 3 x - 1}

f (t) = 3 t^{3} - \frac{2}{t ^{2}}

x \to 0 + lim (\frac{1}{2 x ^{2}})

3 + 5 x

\frac{d}{d x} (x + y z)

What is the derivative of f(x)=ln(x^3-5x^2+3x-1) ?
The derivative of f(x)=ln(x^3-5x^2+3x-1) is (3x^2-10x+3)/(x^3-5x^2+3x-1)
What is the first derivative of f(x)=ln(x^3-5x^2+3x-1) ?
The first derivative of f(x)=ln(x^3-5x^2+3x-1) is (3x^2-10x+3)/(x^3-5x^2+3x-1)