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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\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}

\int \frac{3}{3 x} d x

s im pl i f y 4 sin (x + \frac{π}{3}) - 1

\frac{\partial}{\partial y} (1 + 2 x y)

d er i v a t i v e 1 - e^{3 x^{2}}

\frac{\partial}{\partial x} (3 x^{3} + y^{3})

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