Question

How do you take the derivative of `tan^-1 2x`?

Answer 1

Use the derivative of `tan^-1` and the chain rule.

Explanation:

The derivative of `tan^-1x` is `1/(1+x^2)` (for "why", see note below)

So, applying the chain rule, we get:

`d/dx(tan^-1u) = 1/(1+u^2)*(du)/dx`

In this question `u = 2x`, so we get:

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

`= 2/(1+4x^2)`

Note

If `y = tan^-1x`, then `tany = x`

Differentiating implicitly gets us:

`sec^2y dy/dx = 1," "` so

`dy/dx = 1/sec^2y`

From trigonometry, we know that `1+tan^2y = sec^2y`

so `dy/dx = 1/(1+tan^2y)`

and we have `tany = x`, so we get:

For `y = tan^-1x`, the derivative is:

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