Question

How do you use the chain rule to differentiate `y=sin^5x`?

Answer 1

The chain rule is `d[f(u(x))]/(dx) = (df)/(du)(du)/(dx)` so let `u(x) = sin(x)` then `f(u) = u^5` and it becomes quite simple.

Explanation:

`(df)/(du) = 5u^4`

`(du)/(dx) = cos(x)`

`(df)/(du)(du)/(dx) = 5u^4cos(x)`

Reverse the u substitution:

`(df)/(du)(du)/(dx) = 5sin^4(x)cos(x)`