What is the derivative of log_{7}(e^{-x}cos(pix)) ?

The derivative of log_{7}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(ln(7)cos(pix))

What is the first derivative of log_{7}(e^{-x}cos(pix)) ?

The first derivative of log_{7}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(ln(7)cos(pix))

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\frac{d}{d x} (lo g_{7} (e^{- x} cos (π x)))

- \frac{cos ( π x ) + π sin ( π x )}{ln ( 7 ) cos ( π x )}

Solution steps

\frac{d}{d x} (lo g_{7} (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 ( 7 )})

Take the constant out:

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

= \frac{1}{ln ( 7 )} \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 ( 7 )} \cdot \frac{1}{e ^{- x} cos ( π x )} (- e^{- x} cos (π x) - π e^{- x} sin (π x))

Simplify

\frac{1}{ln ( 7 )} \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 )}{ln ( 7 ) cos ( π x )}

= - \frac{cos ( π x ) + π sin ( π x )}{ln ( 7 ) cos ( π x )}

y = \frac{1}{x + 1}, (1, \frac{1}{2})

(1 + t)^{3}

\int 2 e^{2 t} d t

\frac{d}{d x} ((x^{2} + 2 x)^{4} (x^{2} + 3 x^{2} + 7)^{3})

y = x^{2} - 5 x + 3, (2, - 3)

What is the derivative of log_{7}(e^{-x}cos(pix)) ?
The derivative of log_{7}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(ln(7)cos(pix))
What is the first derivative of log_{7}(e^{-x}cos(pix)) ?
The first derivative of log_{7}(e^{-x}cos(pix)) is -(cos(pix)+pisin(pix))/(ln(7)cos(pix))