Question

How do you find the derivative of `arctan(e^x)`?

Answer 1

`d/dx arctan(e^x)= (e^x)/(e^(2x)+1)`

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(e^x) <=> tany=e^x`

Differentiate Implicitly:

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

Using the `tan"/"sec` identity;

`tan^2y+1 -= sec^2y`
`:. (e^x)^2+1=sec^2y`
`:. e^(2x)+1=sec^2y`

Substituting into [1]

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

Answer 2

`d/dxarctan(e^x)=e^x/(e^(2x)+1)`

Explanation:

Using implicit differentiation together with the known derivatives `d/dx tan(x) = sec^2(x)` and `d/dx e^x = e^x`,

let `y = arctan(e^x)`

`=> tan(y) = tan(arctan(e^x)) = e^x`

`=> d/dxtan(y) = d/dxe^x`

`=> sec^2(y)dy/dx = e^x`

`=> dy/dx = e^x/sec^2(y)`

If we draw a right triangle with an angle `y` and legs such that `tan(y) = e^x`, then we find that `sec(y) = sqrt(e^(2x)+1)`. Using that, we get our final result:

`d/dxarctan(e^x) = dy/dx`

`=e^x/sec^2(y)`

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