What is the derivative of ln(e^x+x) ?

The derivative of ln(e^x+x) is (e^x+1)/(e^x+x)

What is the first derivative of ln(e^x+x) ?

The first derivative of ln(e^x+x) is (e^x+1)/(e^x+x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

derivative of

ln (e^{x} + x)

\frac{e ^{x} + 1}{e ^{x} + x}

Solution steps

\frac{d}{d x} (ln (e^{x} + x))

Apply the chain rule:

\frac{1}{e ^{x} + x} \frac{d}{d x} (e^{x} + x)

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

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

= \frac{1}{e ^{x} + x} (e^{x} + 1)

Simplify

\frac{1}{e ^{x} + x} (e^{x} + 1) : \frac{e ^{x} + 1}{e ^{x} + x}

= \frac{e ^{x} + 1}{e ^{x} + x}

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