What is the derivative of (cos(2x)/(sin(4x))) ?

The derivative of (cos(2x)/(sin(4x))) is -2sin(2x)csc(4x)-4cot(4x)csc(4x)cos(2x)

What is the first derivative of (cos(2x)/(sin(4x))) ?

The first derivative of (cos(2x)/(sin(4x))) is -2sin(2x)csc(4x)-4cot(4x)csc(4x)cos(2x)

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\frac{d}{d x} (\frac{cos ( 2 x )}{sin ( 4 x )})

- 2 sin (2 x) csc (4 x) - 4 cot (4 x) csc (4 x) cos (2 x)

Solution steps

\frac{d}{d x} (\frac{cos ( 2 x )}{sin ( 4 x )})

Simplify

\frac{cos ( 2 x )}{sin ( 4 x )} : cos (2 x) csc (4 x)

= \frac{d}{d x} (cos (2 x) csc (4 x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = cos (2 x), g = csc (4 x)

= \frac{d}{d x} (cos (2 x)) csc (4 x) + \frac{d}{d x} (csc (4 x)) cos (2 x)

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

\frac{d}{d x} (csc (4 x)) = - cot (4 x) csc (4 x) \cdot 4

= (- sin (2 x) \cdot 2) csc (4 x) + (- cot (4 x) csc (4 x) \cdot 4) cos (2 x)

Simplify

= - 2 sin (2 x) csc (4 x) - 4 cot (4 x) csc (4 x) cos (2 x)

f (x) = x^{3} + x^{2} + x + 5

x \to 0 - lim (\frac{1}{1 - 2 ^{\frac{1}{x}}})

\int 9 x - x d x

n = 0 \sum \infty n (\frac{1}{e})^{n}

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

What is the derivative of (cos(2x)/(sin(4x))) ?
The derivative of (cos(2x)/(sin(4x))) is -2sin(2x)csc(4x)-4cot(4x)csc(4x)cos(2x)
What is the first derivative of (cos(2x)/(sin(4x))) ?
The first derivative of (cos(2x)/(sin(4x))) is -2sin(2x)csc(4x)-4cot(4x)csc(4x)cos(2x)