Question

How do you differentiate `f(x)=(x-sin3x)^3/9` using the chain rule?

Answer 1

`1/3*(1-3cos3x)*(x-sin3x)^2`

Explanation:

`d/dy(((x-sin3x)^3)/9)`

= `d/dy(1/9(x-sin3x)^3)`

= `1/9*3*(1-cos3x*3)*(x-sin3x)^2`

= `1/3*(1-3cos3x)*(x-sin3x)^2`