# Differentiate $f(x)=\sin(\sin(\sin(…(\sin(x))))), \text{the number of$\sin$'s is$n$}$.

Determine if $$f(x)$$ has a derivative and find it. $$f(x)$$ is defined as: $$f(x)=\sin(\sin(\sin(...(\sin(x))))), \text{the number of sin's is n}$$

Since $$f(x)$$ is a composition of trigonometric functions which are differentiable, it means that $$f(x)$$ is differentiable. As for determining what the derivative would be, my first guess was using substitution and make $$u$$ the argument of the first "$$\sin$$" and then use the chain rule to differentiate $$\sin(u)$$. However, I am not sure what that would look like here since there are $$n-$$many sinus functions.

• why don't you try $n=2,3$ and see the pattern (it is a product of $n$ terms starting with cosine and then having only sines inside from $n-1$ to $0$ of them) – Conrad Jan 23 at 19:27

Hint : Show inductively that $$f'(x)=\left[\cos(x)\right]\times\left[\cos(\sin(x))\right]\times\left[\cos(\sin(\sin(x))\right]\times\left[...\right]\times\left[\cos(\sin(\sin(\sin(...(x)))))\right]$$
Denoting functional composition by $$\circ$$, we want to differentiate $$\sin^{\circ n}x$$. We prove by induction$$\frac{d}{dx}\sin^{\circ n}x=\prod_{j=0}^{n-1}\cos(\sin^{\circ j}x).$$For $$n=0$$, this is the trivial $$\frac{d}{dx}x=1$$, the RHS being an empty product. If the case $$n=k$$ works, by the chain rule\begin{align}\frac{d}{dx}\sin^{\circ(k+1)}x&=\frac{d}{dx}\sin(\sin^{\circ k}x)\\&=\cos(\sin^{\circ k}x)\frac{d}{dx}\sin^{\circ k}x\\&=\cos(\sin^{\circ k}x)\prod_{j=0}^{k-1}\cos(\sin^{\circ j}x)\\&=\prod_{j=0}^k\cos(\sin^{\circ j}x).\end{align}
Edit: @Sebastiano has requested the $$n=3$$ case explicitly, which requires the previous ones. Continuing from $$n=0$$ which I did as the base step, the $$n=1$$ case is $$\frac{d}{dx}\sin x=\cos x$$. Thereafter:\begin{align}\frac{d}{dx}\sin(\sin x)&=\cos(\sin x)\frac{d}{dx}\sin x\\&=\cos(\sin x)\cos x,\\\frac{d}{dx}\sin(\sin(\sin x))&=\cos(\sin(\sin x))\frac{d}{dx}(\sin(\sin x))\\&=\cos(\sin(\sin x))\cos(\sin x)\cos x.\end{align}