Given $\phi (u)=\sin u, u\in E$, and $f(t)=(t^2,\sqrt{t})$, find an interval $E$ such that $f\circ \phi$ is defined and differentiable on $E$

Question

I need help with this problem:

Given that $\phi (u)=\sin u, u\in E\subset\mathbb{R}$, and $f(t)=(t^2,\sqrt{t}), t\geq 0$, find a suitable interval E such that $f\circ \phi$ is defined and differentiable on E. Apply the Chain Rule to calculate $(f\circ \phi)'(u), u\in E$.

So, I first tried to find $f\circ\phi$ and ended up with $(f\circ\phi)(u)=(\sin^2 u,\sqrt{\sin u})$, am I correct? After that, I don't know how to find the interval $E$ that would make this function defined and differentiable on $E$. How do I find such interval?

Answer 1

For $(f\circ\phi)(u)=(\sin(u)^2,\sqrt{\sin(u)})$ to be defined, both $(\sin(u))^2$ and $\sqrt{\sin(u)}$ need to be defined. The problem is of course with $\sqrt{\sin(u)}$, which is defined if and only if $\sin(u)\geq0$. Can you find an interval on which $\sin(u)\geq0$? Note that there is more than one correct answer.