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

The derivative of 2/(sin^2(2x)) is -8csc^2(2x)cot(2x)

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

The first derivative of 2/(sin^2(2x)) is -8csc^2(2x)cot(2x)

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\frac{d}{d x} (\frac{2}{sin ^{2} ( 2 x )})

- 8 csc^{2} (2 x) cot (2 x)

Solution steps

\frac{d}{d x} (\frac{2}{sin ^{2} ( 2 x )})

Simplify

\frac{2}{sin ^{2} ( 2 x )} : 2 csc^{2} (2 x)

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

Take the constant out:

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

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

Apply the chain rule:

2 csc (2 x) \frac{d}{d x} (csc (2 x))

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

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

= 2 \cdot 2 csc (2 x) (- cot (2 x) csc (2 x) \cdot 2)

Simplify

2 \cdot 2 csc (2 x) (- cot (2 x) csc (2 x) \cdot 2) : - 8 csc^{2} (2 x) cot (2 x)

= - 8 csc^{2} (2 x) cot (2 x)

\int_{1}^{\infty} 6 e^{- 2 x} d x

\int_{0}^{1} x^{2} (x + 1)^{5} d x

\int_{- 32}^{2} \frac{1}{∣ 2 x ∣} d x

y^{^{'}} = \frac{( y - x y )}{x ^{2}}

ln (3 x)

What is the derivative of 2/(sin^2(2x)) ?
The derivative of 2/(sin^2(2x)) is -8csc^2(2x)cot(2x)
What is the first derivative of 2/(sin^2(2x)) ?
The first derivative of 2/(sin^2(2x)) is -8csc^2(2x)cot(2x)