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

The derivative of 1/x ln(e^x+x) is -(ln(e^x+x))/(x^2)+(e^x+1)/(x(e^x+x))

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

The first derivative of 1/x ln(e^x+x) is -(ln(e^x+x))/(x^2)+(e^x+1)/(x(e^x+x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{x}, g = ln (e^{x} + x)

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

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

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

= (- \frac{1}{x ^{2}}) ln (e^{x} + x) + \frac{e ^{x} + 1}{e ^{x} + x} \cdot \frac{1}{x}

Simplify

(- \frac{1}{x ^{2}}) ln (e^{x} + x) + \frac{e ^{x} + 1}{e ^{x} + x} \cdot \frac{1}{x} : - \frac{ln ( e ^{x} + x )}{x ^{2}} + \frac{e ^{x} + 1}{x ( e ^{x} + x )}

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

\frac{\partial}{\partial x} (sin (8 x) cos (2 y))

\int_{1}^{4} x \cdot ln (x) d x

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

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'}} = 6 x

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