What is the (1/(x+1))^' ?

The (1/(x+1))^' is -1/((x+1)^2)

What is the first (1/(x+1))^' ?

The first (1/(x+1))^' is -1/((x+1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\frac{1}{x + 1})^{^{'}}

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

Solution steps

(\frac{1}{x + 1})^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= ((x + 1)^{- 1})^{^{'}}

Apply the chain rule:

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

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

(x + 1)^{'} = 1

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

Simplify

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

\int (2 x - cos (6 x)) d x

x \to \infty lim (0. 8^{x})

\int 3 x^{2} 2 x^{3} - 5 d x

f (x) = 4 - x^{2}, at x = 1.154

x^{3} - 3 x^{2} + 2 x - 2, at x = 0