What is the (arcsin((2x)/(1+x^2)))^' ?

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

What is the first (arcsin((2x)/(1+x^2)))^' ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x^{4} + x^{3} + 2

\frac{\partial}{\partial x} (sin (y) cos (x))

- 8 x^{2} + 5

\int \frac{4 t}{1 - 16 t ^{4}} d t

\int sin (t) sec^{5} (t) d t

What is the (arcsin((2x)/(1+x^2)))^' ?
The (arcsin((2x)/(1+x^2)))^' is -2/(x^2+1)
What is the first (arcsin((2x)/(1+x^2)))^' ?
The first (arcsin((2x)/(1+x^2)))^' is -2/(x^2+1)