Question

How do I find the derivative of `y = arc cot(x) + (cot x)^-1`?

Answer 1

`y^' = sec^2x - 1/(1 + x^2)`

Explanation:

I'll assume that you don't know what the derivative of `"arccot"(x)` is, so that we can use implicit differentiation to find `(dy)/dx`.

First, start by rewriting your function as

`y = "arccot"(x) + 1/cotx" "`, which is equivalent to

`y = "arccot"(x) + tan(x)`

Rerrange to get `arccos(x)` alone on the right-hand side of the equation

`y - tan(x) = "arccot"(x)`

This is equivalent to

`cot(y -tan(x)) = x " "color(blue)((1))`

Differentiate both sides with respect to `x` to get

`d/dxcot(y - tan(x)) = d/dx(x)`

`-csc^2(y - tan(x)) * [(dy)/dx - sec^2(x)] = 1`

`-csc^2(y - tan(x)) * (dy)/dx + csc^2(y - tan(x)) * sec^2(x) = 1`

Rearrange to get `(dy)/dx` alone on the left-hand side of the equation

`-csc^2(y - tan(x)) * (dy)/dx = 1 - csc^2(y - tan(x)) * sec^2(x)`

`(dy)/dx = (1 - csc^2(y - tan(x)) * sec^2(x))/(-csc^2(y - tan(x)))`

`(dy)/dx = -1/csc^2(y - tan(x)) + (color(red)(cancel(color(black)(-csc^2(y - tan(x))))) * sec^2(x))/color(red)(cancel(color(black)(-csc^2(y - tan(x))))`

`(dy)/dx = sec^2x - 1/csc^2(y - tan(x))`

Use the trigonometric identity

`color(blue)(csc^2x = 1 + cot^2x)`

to write

`(dy)/dx = sec^2x - 1/(1 + cot^2(y - tan(x)))`

You know from equation `color(blue)((1))` that

`cot(y - tan(x)) = x`

which means that the derivative will be

`(dy)/dx = color(green)(sec^2x - 1/(1 + x^2))`