What is the derivative of f(x)=(x+1)(-1/(x+2)) ?

The derivative of f(x)=(x+1)(-1/(x+2)) is -1/((x+2)^2)

What is the first derivative of f(x)=(x+1)(-1/(x+2)) ?

The first derivative of f(x)=(x+1)(-1/(x+2)) is -1/((x+2)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = (x + 1) (- \frac{1}{x + 2})

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

Solution steps

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

Simplify

(x + 1) (- \frac{1}{x + 2}) : - \frac{x + 1}{x + 2}

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (3 arctan (x y^{2}))

(cos (θ))^{^{'}}

\int_{- 1}^{3} (u^{3} + u) d u

y = x^{2} - 5 x

\frac{4}{s ( s + 2 ) ^{2}}

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