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

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

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

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

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

- 1 + cot (x)

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= - 1 + cot (x)

\int 1 + (tan (x))^{2} d x

g (x) = 6 - x

\frac{1}{1 - x}, 0, 1

\int 3 x^{5} d x

n = 9 \sum \infty e^{3 - 4 n}

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