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

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

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

The first derivative of (2cos(x)/(sec(x))) is -2sin(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} (\frac{2 cos ( x )}{sec ( x )})

- 2 sin (2 x)

Solution steps

\frac{d}{d x} (\frac{2 cos ( x )}{sec ( x )})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{cos ( x )}{sec ( x )})

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= - 2 sin (2 x)

y^{^{''}} + y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 0

(2 x y) d y + (x^{2} + y^{2}) d x = 0

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y = (tan (x))^{\frac{9}{x}}

\int (x^{2} x) d x

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