Question

Question #3a400

Answer 1

`y'=sec(x)^2*tan(x)^(-2/3)/3-2/3*sec(2x)^2*tan(2x)^(-2/3)`

Explanation:

Supposing you mean

`y=root(3)(tan(x))-root(3)(tan(2x))`

and you are looking for `y'`

`y=color(red)(tan)(color(green)(x))^(color(blue)(1/3))-color(red)(tan)(color(green)(2x))^color(blue)(1/3)`

`y'=color(blue)(1/3)*color(green)(1)*color(red)(sec(x)^2)*color(red)(tan)(color(green)(x))^color(blue)(-2/3)-color(blue)(1/3)*color(green)(2)*color(red)(sec(2x)^2)*color(red)(tan)(color(green)(2x))^color(blue)(-2/3)`
`y'=sec(x)^2*tan(x)^(-2/3)/3-2/3*sec(2x)^2*tan(2x)^(-2/3)`