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

The derivative of log_{e}(sin(2x)+cot(2x)) is (2cos(2x)-2csc^2(2x))/(sin(2x)+cot(2x))

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

The first derivative of log_{e}(sin(2x)+cot(2x)) is (2cos(2x)-2csc^2(2x))/(sin(2x)+cot(2x))

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derivative of

lo g_{e} (sin (2 x) + cot (2 x))

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

Solution steps

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

Simplify

lo g_{e} (sin (2 x) + cot (2 x)) : ln (sin (2 x) + cot (2 x))

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

Apply the chain rule:

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

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

\frac{d}{d x} (sin (2 x) + cot (2 x)) = cos (2 x) \cdot 2 - csc^{2} (2 x) \cdot 2

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

Simplify

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

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

y = \frac{1}{2} x

\int_{π}^{2 π} (t - π) e^{- s t} d t

x y^{^{'}} - 2 y = x^{5}

13 - x^{2}

\frac{\partial}{\partial y} (3 arctan (\frac{x}{y}))

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