What is the derivative of g(x)=ln(xe^{-2x}) ?

The derivative of g(x)=ln(xe^{-2x}) is 1/x-2

What is the first derivative of g(x)=ln(xe^{-2x}) ?

The first derivative of g(x)=ln(xe^{-2x}) is 1/x-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

g (x) = ln (x e^{- 2 x})

\frac{1}{x} - 2

Solution steps

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

Simplify

ln (x e^{- 2 x}) : ln (x) - 2 x

= \frac{d}{d x} (ln (x) - 2 x)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (ln (x)) - \frac{d}{d x} (2 x)

\frac{d}{d x} (ln (x)) = \frac{1}{x}

\frac{d}{d x} (2 x) = 2

= \frac{1}{x} - 2

\int \frac{x ^{4} + 10 x ^{3}}{2 x ^{2}} d x

\int 2 (1 - 3 x)^{2} d x

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

\int \frac{1}{e ^{- x} + e ^{x}} d x

t \to 0 + lim (\frac{4}{t} - \frac{1}{e ^{t} - 1})

What is the derivative of g(x)=ln(xe^{-2x}) ?
The derivative of g(x)=ln(xe^{-2x}) is 1/x-2
What is the first derivative of g(x)=ln(xe^{-2x}) ?
The first derivative of g(x)=ln(xe^{-2x}) is 1/x-2