What is the derivative of-4cot(2xcsc(2x)) ?

The derivative of-4cot(2xcsc(2x)) is -4(-2csc^3(2x)-2cot^2(2x)csc(2x))

What is the first derivative of-4cot(2xcsc(2x)) ?

The first derivative of-4cot(2xcsc(2x)) is -4(-2csc^3(2x)-2cot^2(2x)csc(2x))

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\frac{d}{d x} (- 4 cot (2 x) csc (2 x))

- 4 (- 2 csc^{3} (2 x) - 2 cot^{2} (2 x) csc (2 x))

Solution steps

\frac{d}{d x} (- 4 cot (2 x) csc (2 x))

Take the constant out:

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

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

Apply the Product Rule:

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

f = cot (2 x), g = csc (2 x)

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

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

\frac{d}{d x} (csc (2 x)) = - cot (2 x) csc (2 x) \cdot 2

= - 4 ((- csc^{2} (2 x) \cdot 2) csc (2 x) + (- cot (2 x) csc (2 x) \cdot 2) cot (2 x))

Simplify

- 4 ((- csc^{2} (2 x) \cdot 2) csc (2 x) + (- cot (2 x) csc (2 x) \cdot 2) cot (2 x)) : - 4 (- 2 csc^{3} (2 x) - 2 cot^{2} (2 x) csc (2 x))

= - 4 (- 2 csc^{3} (2 x) - 2 cot^{2} (2 x) csc (2 x))

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 sin (2 t)

\frac{d}{d x} (\frac{1}{3} cos (\frac{6 π x - 1}{2}))

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

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

y = 2 x^{2}

What is the derivative of-4cot(2xcsc(2x)) ?
The derivative of-4cot(2xcsc(2x)) is -4(-2csc^3(2x)-2cot^2(2x)csc(2x))
What is the first derivative of-4cot(2xcsc(2x)) ?
The first derivative of-4cot(2xcsc(2x)) is -4(-2csc^3(2x)-2cot^2(2x)csc(2x))