`f(t)=arctan(sinh(t))`

Take note that the derivative formula of arctangent is

- `d/dx[arctan(u)]=1/(1+u^2)*(du)/dx`

Applying this, the derivative of the function will be

`f'(t) = d/(dt)[arctan(sinh(t))]`

`f'(t) = 1/(1+sinh^2(t)) *d/(dt)[sinh(t)]`

Also, the derivative formula of hyperbolic sine is

- `d/dx[sinh(u)]=cosh(u)*(du)/(dx)`

Applying this, f'(t) will become

`f'(t)= 1/(1+sinh^2(t)) *cosh(t)*d/(dt)(t)`

`f'(t)= 1/(1+sinh^2(t)) *cosh(t)*1`

`f'(t)= cosh(t)/(1+sinh^2(t))`

`f'(t)=...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

`f(t)=arctan(sinh(t))`

Take note that the derivative formula of arctangent is

- `d/dx[arctan(u)]=1/(1+u^2)*(du)/dx`

Applying this, the derivative of the function will be

`f'(t) = d/(dt)[arctan(sinh(t))]`

`f'(t) = 1/(1+sinh^2(t)) *d/(dt)[sinh(t)]`

Also, the derivative formula of hyperbolic sine is

- `d/dx[sinh(u)]=cosh(u)*(du)/(dx)`

Applying this, f'(t) will become

`f'(t)= 1/(1+sinh^2(t)) *cosh(t)*d/(dt)(t)`

`f'(t)= 1/(1+sinh^2(t)) *cosh(t)*1`

`f'(t)= cosh(t)/(1+sinh^2(t))`

`f'(t)= cosh(t)/(cosh^2(t))`

`f'(t)= 1/cosh(t)` is the final derivative