how do u determine this derivative using the chain rule?

Question

y = 5 cot (2/x)

Answer 1

Simply work from the outside in:

dy/dx
d(5 cot (2/x))/dx
5 d(cot (2/x))/dx
-5 csc² (2/x) d(2/x)/dx
-5 csc² (2/x) (-2/x²)
10 csc² (2/x)/x²

And we are done.

Answer 2

chain rule:

d/dx g ( f(x) ) = g' ( f(x) ) * f'(x)

y = 5 cot (2/x)

let u be 2/x
then dy/du cot(u) = -csc^2(u)

d/dx (2/x) = -2/x^2

y' = 5 * -csc^2(u) * -2/x^2

recall that u = 2/x

y' = 5 * -csc^2(x/2) * -2/x^2

simplify
y' = 10 csc^2(x/2) / x^2 <== answer

Answer 3

y = 2/x ->der(5 cot(y)) = -5(csc(y))^2 chain rule der(2/x) = -2/x^2

so (10(csc(2/x))^2) / x^2

Answer 4

(dy/du)(du/dx) = dy/dx
{u = 2/x
{du = -2/x^2 dx
y = 5 cot(2/x)
dy/du = 5 cot(u) d/du
dy/du = 5 cot(u) d/du
(5 cot(u) d/du)(-2/x^2)
Plug for u