What is the derivative of ln(2e^{-t}sin(t)) ?

The derivative of ln(2e^{-t}sin(t)) is -1+cot(t)

What is the first derivative of ln(2e^{-t}sin(t)) ?

The first derivative of ln(2e^{-t}sin(t)) is -1+cot(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

ln (2 e^{- t} sin (t))

- 1 + cot (t)

Solution steps

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

Apply the chain rule:

\frac{1}{2 e ^{- t} sin ( t )} \frac{d}{d t} (2 e^{- t} sin (t))

= \frac{1}{2 e ^{- t} sin ( t )} \frac{d}{d t} (2 e^{- t} sin (t))

\frac{d}{d t} (2 e^{- t} sin (t)) = 2 (- e^{- t} sin (t) + e^{- t} cos (t))

= \frac{1}{2 e ^{- t} sin ( t )} \cdot 2 (- e^{- t} sin (t) + e^{- t} cos (t))

Simplify

\frac{1}{2 e ^{- t} sin ( t )} \cdot 2 (- e^{- t} sin (t) + e^{- t} cos (t)) : - 1 + cot (t)

= - 1 + cot (t)

f (x) = x^{2} + 3 x + 5

\frac{\partial}{\partial y} (e^{- y})

x \to 1 lim (2 x - 2)

y = \frac{3}{x}

What is the derivative of ln(2e^{-t}sin(t)) ?
The derivative of ln(2e^{-t}sin(t)) is -1+cot(t)
What is the first derivative of ln(2e^{-t}sin(t)) ?
The first derivative of ln(2e^{-t}sin(t)) is -1+cot(t)