Question

How do you find the derivative `f(x)=arctan(x/alpha)`?

Answer 1

`f'(x)=alpha/(x^2+alpha^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.

`y=arctan(x/alpha) <=> tany=x/alpha`

Differentiate Implicitly:

`sec^2ydy/dx = 1/alpha` ..... [1]

Using the `tan"/"sec` identity;

`tan^2y+1 = sec^2y`
`:. (x/alpha)^2+1=sec^2y`

Substituting into [1]

`((x/alpha)^2+1)dy/dx=1/alpha`
`:. ((x^2+alpha^2)/alpha^2)dy/dx=1/alpha`
`:. dy/dx=1/alpha * alpha^2/(x^2+alpha^2)`
`:. dy/dx=alpha/(x^2+alpha^2)`