Question

What is the derivative of `arctan(x-1)`?

Answer 1

`d/dx arctan(x-1) = 1/(x^2-2x+2)`

Explanation:

We can differentiate this using implicit differentiation:

Let `y = arctan(x-1)`

`=> tan(y) = x-1`

`=> d/dxtan(y) = d/dx(x-1)`

`=> sec^2(y)dy/dx = 1`

`=> dy/dx = 1/sec^2(y)`

Drawing a right triangle with an angle `y` such that `tan(y) = x-1`, we can find that `sec(y) = sqrt(x^2-2x+2)`. Plugging that in, we get our result:

`d/dx arctan(x-1) = 1/(x^2-2x+2)`

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Note that the same process shows that the formula for `d/dxarctan(x) = 1/(x^2+1)`. If we already have that formula, then we can simply use the chain rule:

`d/dxarctan(x-1) = 1/((x-1)^2+1)(d/dx(x-1))`

`=1/(x^2-2x+1+1)*1`
`=1/(x^2-2x+2)`