What is the derivative of log_{4}(e^{2x}cos(3x)) ?

The derivative of log_{4}(e^{2x}cos(3x)) is (2cos(3x)-3sin(3x))/(2ln(2)cos(3x))

What is the first derivative of log_{4}(e^{2x}cos(3x)) ?

The first derivative of log_{4}(e^{2x}cos(3x)) is (2cos(3x)-3sin(3x))/(2ln(2)cos(3x))

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derivative of

lo g_{4} (e^{2 x} cos (3 x))

\frac{2 cos ( 3 x ) - 3 sin ( 3 x )}{2 ln ( 2 ) cos ( 3 x )}

Solution steps

\frac{d}{d x} (lo g_{4} (e^{2 x} cos (3 x)))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( e ^{2 x} cos ( 3 x ) )}{ln ( 4 )})

Take the constant out:

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

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (e^{2 x} cos (3 x)))

Apply the chain rule:

\frac{1}{e ^{2 x} cos ( 3 x )} \frac{d}{d x} (e^{2 x} cos (3 x))

= \frac{1}{e ^{2 x} cos ( 3 x )} \frac{d}{d x} (e^{2 x} cos (3 x))

\frac{d}{d x} (e^{2 x} cos (3 x)) = 2 e^{2 x} cos (3 x) - 3 e^{2 x} sin (3 x)

= \frac{1}{ln ( 4 )} \cdot \frac{1}{e ^{2 x} cos ( 3 x )} (2 e^{2 x} cos (3 x) - 3 e^{2 x} sin (3 x))

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{e ^{2 x} cos ( 3 x )} (2 e^{2 x} cos (3 x) - 3 e^{2 x} sin (3 x)) : \frac{2 cos ( 3 x ) - 3 sin ( 3 x )}{2 ln ( 2 ) cos ( 3 x )}

= \frac{2 cos ( 3 x ) - 3 sin ( 3 x )}{2 ln ( 2 ) cos ( 3 x )}

\frac{30 x ^{2}}{( - 5 x ^{3} + 4 ) ^{3}}

x^{2} + x y - y^{2} = - 11, (1, 4)

t^{2} y^{^{''}} + 3 t y^{^{'}} - 3 y = \frac{1}{t}

\frac{2}{x - 3}

\int cos (x) (7 + 8 sin^{2} (x)) d x

What is the derivative of log_{4}(e^{2x}cos(3x)) ?
The derivative of log_{4}(e^{2x}cos(3x)) is (2cos(3x)-3sin(3x))/(2ln(2)cos(3x))
What is the first derivative of log_{4}(e^{2x}cos(3x)) ?
The first derivative of log_{4}(e^{2x}cos(3x)) is (2cos(3x)-3sin(3x))/(2ln(2)cos(3x))