Question

Question #087c5

Answer 1

`{dy}/{dx} = 1/{1+x^2}`

Explanation:

Let `y = cot^{-1}({1+x}/{1-x})`. Then
`{1+x}/{1-x} = cot y = {cos y}/{sin y}`
which can be easily solved to yield

`x = {cos y+ sin y}/{cos y-sin y}= tan(pi/4+y)`

Differentiating both sides with respect to `x` yields

`1 = sec^2(pi/4+y) {dy}/{dx} = (1+x^2) {dy}/{dx}`

Thus

`{dy}/{dx} = 1/{1+x^2}`