What is the derivative of-3cot(2xsec(2x)) ?

The derivative of-3cot(2xsec(2x)) is -3(-2csc^2(2x)sec(2x)+2sec(2x)tan(2x)cot(2x))

What is the first derivative of-3cot(2xsec(2x)) ?

The first derivative of-3cot(2xsec(2x)) is -3(-2csc^2(2x)sec(2x)+2sec(2x)tan(2x)cot(2x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (- 3 cot (2 x) sec (2 x))

- 3 (- 2 csc^{2} (2 x) sec (2 x) + 2 sec (2 x) tan (2 x) cot (2 x))

Solution steps

\frac{d}{d x} (- 3 cot (2 x) sec (2 x))

Take the constant out:

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

= - 3 \frac{d}{d x} (cot (2 x) sec (2 x))

Apply the Product Rule:

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

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

= - 3 (\frac{d}{d x} (cot (2 x)) sec (2 x) + \frac{d}{d x} (sec (2 x)) cot (2 x))

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

\frac{d}{d x} (sec (2 x)) = sec (2 x) tan (2 x) \cdot 2

= - 3 ((- csc^{2} (2 x) \cdot 2) sec (2 x) + sec (2 x) tan (2 x) \cdot 2 cot (2 x))

Simplify

= - 3 (- 2 csc^{2} (2 x) sec (2 x) + 2 sec (2 x) tan (2 x) cot (2 x))

\int x^{2} + 8 e^{x} + C d x

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x} + e^{2 x}

\int \frac{1}{3 x + 2} d x

\int_{0}^{2} (x - 3)^{2} - (x - 1)^{2} d x

(x + 1) y^{^{'}} + x y = 0

What is the derivative of-3cot(2xsec(2x)) ?
The derivative of-3cot(2xsec(2x)) is -3(-2csc^2(2x)sec(2x)+2sec(2x)tan(2x)cot(2x))
What is the first derivative of-3cot(2xsec(2x)) ?
The first derivative of-3cot(2xsec(2x)) is -3(-2csc^2(2x)sec(2x)+2sec(2x)tan(2x)cot(2x))