Question

How do you differentiate `f(x)=1/sin(e^arctanx)` using the chain rule?

Answer 1

`dy/dx = -(csc(e^(arctan(x)))cot(e^(arctan(x)))e^(arctan(x)))/(x^2+1)`

Explanation:

So we have

`y = 1/sin(e^(arctan(x))) = csc(e^(arctan(x)))`

While the cosecant has a tabled, standard derivative we can work with the sine too, if you don't remember / don't have a standard derivative table handy.

Let's say that `u = sin(e^(arctan(x)))`

`dy/dx = d/(du)u^(-1)(du)/dx`
`dy/dx = -u^(-3/2)d/dx(sin(e^(arctan(x)))`

Let's say that `v = e^(arctan(x))`

`dy/dx = -u^(-3/2)d/(dv)sin(v)(dv)/dx = -u^(-3/2)cos(v)d/dx(e^(arctan(x)))`

Let's say that `w = arctan(x)`

`dy/dx = -u^(-3/2)cos(v)d/(dw)e^(w)(dw)/dx`
`dy/dx = -u^(-3/2)cos(v)e^(w)d/dx(arctan(x))`
`dy/dx = -(u^(-3/2)cos(v)e^(w))/(x^2+1)`

Putting it all in terms of `x`

`dy/dx = -(csc(e^(arctan(x)))cot(e^(arctan(x)))e^(arctan(x)))/(x^2+1)`