What is the derivative of (2cos(x)/(x+1)) ?

The derivative of (2cos(x)/(x+1)) is (2(-sin(x)(x+1)-cos(x)))/((x+1)^2)

What is the first derivative of (2cos(x)/(x+1)) ?

The first derivative of (2cos(x)/(x+1)) is (2(-sin(x)(x+1)-cos(x)))/((x+1)^2)

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 )}{x + 1})

\frac{2 ( - sin ( x ) ( x + 1 ) - cos ( x ) )}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 cos ( x )}{x + 1})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{cos ( x )}{x + 1})

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

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

\frac{d}{d x} (x + 1) = 1

= 2 \cdot \frac{( - sin ( x ) ) ( x + 1 ) - 1 \cdot cos ( x )}{( x + 1 ) ^{2}}

Simplify

2 \cdot \frac{( - sin ( x ) ) ( x + 1 ) - 1 \cdot cos ( x )}{( x + 1 ) ^{2}} : \frac{2 ( - sin ( x ) ( x + 1 ) - cos ( x ) )}{( x + 1 ) ^{2}}

= \frac{2 ( - sin ( x ) ( x + 1 ) - cos ( x ) )}{( x + 1 ) ^{2}}

x \to 0 + lim (x)

y^{^{'}} + y = e^{x} sin (e^{x})

y^{^{'}} - (y - 1)^{2} = 0

\int X d X

x \to 0 lim ((1 + \frac{1}{x})^{5 x})

What is the derivative of (2cos(x)/(x+1)) ?
The derivative of (2cos(x)/(x+1)) is (2(-sin(x)(x+1)-cos(x)))/((x+1)^2)
What is the first derivative of (2cos(x)/(x+1)) ?
The first derivative of (2cos(x)/(x+1)) is (2(-sin(x)(x+1)-cos(x)))/((x+1)^2)