Question

How do you differentiate `y=tan^-1(3x)`?

Answer 1

`= 3/(9x^2 + 1)`

Explanation:

for `d/dx (tan^-1(3x))`

you can remember that

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

and that, where `u = u(v)`, via the chain rule:

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

or you can switch the function over by saying that

`tan y = 3x` and then differentiating implicitly, so that

`sec^2 y \ y' = 3`

BTW you are still using the chain rule because:

`(d(tan y))/dx = (d(tan y))/dy * dy/dx = sec^2 y * dy/dx`

All of which means that:

`\ y' = 3/sec^2 y`

`\ y' = 3/(tan^2 y + 1)`

`\ y' = 3/((3x)^2 + 1)`

`= 3/(9x^2 + 1)`