What is the (4arccos(sqrt(1-x^2)))^' ?

The (4arccos(sqrt(1-x^2)))^' is 4/(sqrt(1-x^2))

What is the first (4arccos(sqrt(1-x^2)))^' ?

The first (4arccos(sqrt(1-x^2)))^' is 4/(sqrt(1-x^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(4 arccos (1 - x^{2}))^{^{'}}

\frac{4}{1 - x ^{2}}

Solution steps

(4 arccos (1 - x^{2}))^{'}

Take the constant out:

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

= 4 (arccos (1 - x^{2}))^{^{'}}

Apply the chain rule:

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

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

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

= 4 - \frac{1}{1 - ( 1 - x ^{2} ) ^{2}} (- \frac{x}{1 - x ^{2}})

Simplify

4 - \frac{1}{1 - ( 1 - x ^{2} ) ^{2}} (- \frac{x}{1 - x ^{2}}) : \frac{4}{1 - x ^{2}}

= \frac{4}{1 - x ^{2}}

\frac{d}{d x} (x^{2} + y^{2} - 9)

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

x \to 4 + lim (5)

\int sin (ln (14 x)) d x

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

What is the (4arccos(sqrt(1-x^2)))^' ?
The (4arccos(sqrt(1-x^2)))^' is 4/(sqrt(1-x^2))
What is the first (4arccos(sqrt(1-x^2)))^' ?
The first (4arccos(sqrt(1-x^2)))^' is 4/(sqrt(1-x^2))