What is the derivative of (cos(x)/(-2sin(2x))) ?

The derivative of (cos(x)/(-2sin(2x))) is -(-sin(x)sin(2x)-2cos(2x)cos(x))/(2sin^2(2x))

What is the first derivative of (cos(x)/(-2sin(2x))) ?

The first derivative of (cos(x)/(-2sin(2x))) is -(-sin(x)sin(2x)-2cos(2x)cos(x))/(2sin^2(2x))

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\frac{d}{d x} (\frac{cos ( x )}{- 2 sin ( 2 x )})

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

Solution steps

\frac{d}{d x} (\frac{cos ( x )}{- 2 sin ( 2 x )})

Simplify

\frac{cos ( x )}{- 2 sin ( 2 x )} : - \frac{cos ( x )}{2 sin ( 2 x )}

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

\int \frac{9}{x ^{3}} d x

\frac{\partial}{\partial r} (z)

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

13 x d x = 127 (92) y d y

x^{^{'}} = 1 - x^{2}

What is the derivative of (cos(x)/(-2sin(2x))) ?
The derivative of (cos(x)/(-2sin(2x))) is -(-sin(x)sin(2x)-2cos(2x)cos(x))/(2sin^2(2x))
What is the first derivative of (cos(x)/(-2sin(2x))) ?
The first derivative of (cos(x)/(-2sin(2x))) is -(-sin(x)sin(2x)-2cos(2x)cos(x))/(2sin^2(2x))