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

The derivative of ln(sqrt(e^{-2x)+cos(x})) is (-2e^{-2x}-sin(x))/(2(e^{-2x)+cos(x))}

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

The first derivative of ln(sqrt(e^{-2x)+cos(x})) is (-2e^{-2x}-sin(x))/(2(e^{-2x)+cos(x))}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (sin^{3} (3 x^{2} + 6 x))

\int_{0}^{3} (\frac{x}{3} + 5) d x

\frac{d}{d x} (\frac{5 - x ^{2}}{5 + x ^{2}})

\frac{\partial}{\partial x} (\frac{1}{6} x^{3})

f (x) = 2 x^{2} + 3 x - 2, (- 1, - 3)

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