Question

How do you differentiate `f(x)=e^(csc(2/x)` using the chain rule.?

Answer 1

This will require multiple applications of the chain rule.

Let's start with `csc(2/x)`. Let `y = cscu` and `u = 2/x`.

Let's differentiate:

`y = 1/sinu`

`y' = ((0 xx sinu) - (1 xx cosu))/(sinu)^2`

`y' = -cosu/sin^2u`

`y' = -cotucscu`

`u' = ((0 xx x) - (2 xx 1))/(x)^2`

`u' = -2/x^2`

`dy/dx = dy/(du) xx (du)/dx`

`dy/dx = -2/x^2 xx -cotucscu`

`dy/dx = -(-2cotucscu)/x^2`

`dy/dx = (2cot(2/x)csc(2/x))/x^2`

Now that we know this derivative we can say:

let `y = e^u` and `u = (2cot(2/x)csc(2/x))/x^2`

We already know the derivative of `u` and the derivative of `y` is `e^u`.

Hence:

`f'(x) = e^u xx (2cot(2/x)csc(2/x))/x^2`

`f'(x) = (e^u(2cot(2/x)csc(2/x)))/x^2`

`f'(x) = (e^(csc(2/x))((2cot(2/x)csc(2/x))))/x^2`

Hopefully this helps!