What is the derivative of f(x)=arctan((x-1)/(x+1)) ?

The derivative of f(x)=arctan((x-1)/(x+1)) is 1/(x^2+1)

What is the first derivative of f(x)=arctan((x-1)/(x+1)) ?

The first derivative of f(x)=arctan((x-1)/(x+1)) is 1/(x^2+1)

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derivative of

f (x) = arctan (\frac{x - 1}{x + 1})

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

Solution steps

\frac{d}{d x} (arctan (\frac{x - 1}{x + 1}))

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = x, at x = 4

f (x) = 9 - x^{2}, at x = - 1

x^{- \frac{2}{3}}

\int \frac{7 x + 6}{x ^{2} ( x + 2 )} d x

y = 8 x^{2} + 8, (2, 40)

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