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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x (x - 1)

y^{^{'}} = (x + 1)^{2}

f (x) = 3 x^{3} - 3 x, (1, 0)

\int sin (2) (π x) cos (5) (π x) d x

y^{^{''}} - 2 y^{^{'}} + 1 y = 0

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