What is the derivative of ln(8x^2+1) ?

The derivative of ln(8x^2+1) is (16x)/(8x^2+1)

What is the first derivative of ln(8x^2+1) ?

The first derivative of ln(8x^2+1) is (16x)/(8x^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (ln (8 x^{2} + 1))

\frac{16 x}{8 x ^{2} + 1}

Solution steps

\frac{d}{d x} (ln (8 x^{2} + 1))

Apply the chain rule:

\frac{1}{8 x ^{2} + 1} \frac{d}{d x} (8 x^{2} + 1)

= \frac{1}{8 x ^{2} + 1} \frac{d}{d x} (8 x^{2} + 1)

\frac{d}{d x} (8 x^{2} + 1) = 16 x

= \frac{1}{8 x ^{2} + 1} \cdot 16 x

Simplify

\frac{1}{8 x ^{2} + 1} \cdot 16 x : \frac{16 x}{8 x ^{2} + 1}

= \frac{16 x}{8 x ^{2} + 1}

\int \frac{- 3 x ^{3} + 2 x ^{2} - 3 x - 6}{x ^{3} + x} d x

n = 0 \sum \infty 3^{n} x^{2 n}

\frac{1}{3} x^{3} - \frac{16}{x} - \frac{40}{3}

h \to 0 lim (7)

\int (8 x^{3} - 9 x^{2} + 4) d x

What is the derivative of ln(8x^2+1) ?
The derivative of ln(8x^2+1) is (16x)/(8x^2+1)
What is the first derivative of ln(8x^2+1) ?
The first derivative of ln(8x^2+1) is (16x)/(8x^2+1)