Question

How do you find the derivative of `y=arc cot(x)`?

Answer 1

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

Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you can use the chain rule.

Let `y=arc cot(x) <=> coty=x`

Differentiate Implicitly:

`-csc^2ydy/dx = 1` ..... [1]

Using the `csc"/"cot` identity;

`1+cot^2y=csc^2y`
`:. 1+x^2=csc^2y`
`:. csc^2y=1+x^2`

Substituting into [1]
`:. -(1+x^2)dy/dx=1`
`:. dy/dx = -1/(1+x^2)`