What is the derivative of log_{4}(e^{-x}cos(pix)) ?

The derivative of log_{4}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(2ln(2)cos(pix))

What is the first derivative of log_{4}(e^{-x}cos(pix)) ?

The first derivative of log_{4}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(2ln(2)cos(pix))

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derivative of

lo g_{4} (e^{- x} cos (π x))

- \frac{cos ( π x ) + π sin ( π x )}{2 ln ( 2 ) cos ( π x )}

Solution steps

\frac{d}{d x} (lo g_{4} (e^{- x} cos (π x)))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( e ^{- x} cos ( π x ) )}{ln ( 4 )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (e^{- x} cos (π x)))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 4 )} \cdot \frac{1}{e ^{- x} cos ( π x )} (- e^{- x} cos (π x) - π e^{- x} sin (π x))

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{e ^{- x} cos ( π x )} (- e^{- x} cos (π x) - π e^{- x} sin (π x)) : - \frac{cos ( π x ) + π sin ( π x )}{2 ln ( 2 ) cos ( π x )}

= - \frac{cos ( π x ) + π sin ( π x )}{2 ln ( 2 ) cos ( π x )}

\int 4 cos (θ) sin (θ) d θ

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

\frac{1}{( s + 1 ) ^{\frac{1}{2}}}

(- 6, 4), (- 1, 4)

\frac{d}{d x} (\frac{x ^{n - 1}}{n})

What is the derivative of log_{4}(e^{-x}cos(pix)) ?
The derivative of log_{4}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(2ln(2)cos(pix))
What is the first derivative of log_{4}(e^{-x}cos(pix)) ?
The first derivative of log_{4}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(2ln(2)cos(pix))