Let $f(x)=\dfrac{\sin x}{(1+\sin^n(x))^{1/n}}$ and $g(x)=\underset{n\text{ times}}{\underbrace{f \circ f \circ f \circ f\circ f \circ \cdots\circ f}}(x)$. Where $\circ$ represents function composition.
Show that $g(x)=\dfrac{\sin x}{(1+n\sin^n(x))^{1/n}}$.
I cant spot any pattern in the composition function. Each time it's becoming more complicated. Am I missing something?
Let an $n\in{\mathbb N}_{\geq1}$ be fixed, and consider the function $$F(y):={y\over(1+y^n)^{1/n}}\qquad(y\geq0)\ .$$ This $F$ can be considered as composition $$F={\rm root}_n\circ T\circ {\rm pow}_n\ ,\tag{1}$$ whereby $${\rm root}_n(t):=t^{1/n},\quad {\rm pow}_n(t):=t^n\qquad(t\geq0)\ ,$$ and $$T(t):={t\over 1+t}\ .$$ It is well known, and easily verified by induction that $$T^{\circ n}(t)={t\over 1 +n\> t}\ .$$ As ${\rm pow}_n\circ {\rm root}_n={\rm id}_{{\mathbb R}_{\geq0}}$ we therefore obtain from $(1)$ that $$F^{\circ n}={\rm root}_n\circ T^n\circ{\rm pow}_n\ ,$$ or $$F^{\circ n}(y)=\left({y^n\over 1+ n y^n}\right)^{1/n}={y\over(1+n y^n)^{1/n}}\qquad(y\geq0)\ .$$ Putting $y:=\sin x$ here (assuming $\sin x\geq0$) we therefore have $$F^{\circ n}(\sin x)={\sin x\over(1+n \sin^n x)^{1/n}}\ .\tag{2}$$ This looks promising, but is not what you wanted: Your $f$ is defined by $f=F\circ\sin$, but from this it does not follow that $$f^{\circ n}=F^{\circ n}\circ\sin\ .$$ Instead one has, e.g., $$f^{\circ3}=F\circ\sin\circ F\circ\sin \circ F\circ\sin\ .$$ The relation $(2)$ therefore captures what can be saved from the hoped for identity.