What is the derivative of log_{2}(sin(2x+x)) ?

The derivative of log_{2}(sin(2x+x)) is (2cos(2x)+1)/(ln(2)(sin(2x)+x))

What is the first derivative of log_{2}(sin(2x+x)) ?

The first derivative of log_{2}(sin(2x+x)) is (2cos(2x)+1)/(ln(2)(sin(2x)+x))

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\frac{d}{d x} (lo g_{2} (sin (2 x) + x))

\frac{2 cos ( 2 x ) + 1}{ln ( 2 ) ( sin ( 2 x ) + x )}

Solution steps

\frac{d}{d x} (lo g_{2} (sin (2 x) + x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( sin ( 2 x ) + x )}{ln ( 2 )})

Take the constant out:

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

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (sin (2 x) + x))

Apply the chain rule:

\frac{1}{sin ( 2 x ) + x} \frac{d}{d x} (sin (2 x) + x)

= \frac{1}{sin ( 2 x ) + x} \frac{d}{d x} (sin (2 x) + x)

\frac{d}{d x} (sin (2 x) + x) = cos (2 x) \cdot 2 + 1

= \frac{1}{ln ( 2 )} \cdot \frac{1}{sin ( 2 x ) + x} (cos (2 x) \cdot 2 + 1)

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{sin ( 2 x ) + x} (cos (2 x) \cdot 2 + 1) : \frac{2 cos ( 2 x ) + 1}{ln ( 2 ) ( sin ( 2 x ) + x )}

= \frac{2 cos ( 2 x ) + 1}{ln ( 2 ) ( sin ( 2 x ) + x )}

\frac{d v}{d t} = 0.32 v + 0.22 (100 e^{0.32 t})

\frac{d}{d x} (\frac{x ^{2}}{x - 4})

\int \frac{x}{( x - 6 ) ( x + 2 ) ^{2}} d x

f (x) = 2 - x^{2}, g (x) = x

\int x + 1 d x

What is the derivative of log_{2}(sin(2x+x)) ?
The derivative of log_{2}(sin(2x+x)) is (2cos(2x)+1)/(ln(2)(sin(2x)+x))
What is the first derivative of log_{2}(sin(2x+x)) ?
The first derivative of log_{2}(sin(2x+x)) is (2cos(2x)+1)/(ln(2)(sin(2x)+x))