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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.

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    $\begingroup$ 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) $\endgroup$ – Conrad Jan 23 at 19:27
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Hint : Show inductively that $$f'(x)=\left[\cos(x)\right]\times\left[\cos(\sin(x))\right]\times\left[\cos(\sin(\sin(x))\right]\times\left[...\right]\times\left[\cos(\sin(\sin(\sin(...(x)))))\right]$$

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Just to flesh out @TheSilverDoe's answer:

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:$$\begin{align}\frac{d}{dx}\sin(\sin x)&=\cos(\sin x)\frac{d}{dx}\sin x\\&=\cos(\sin x)\cos x,\\\frac{d}{dx}\sin(\sin(\sin x))&=\cos(\sin(\sin x))\frac{d}{dx}(\sin(\sin x))\\&=\cos(\sin(\sin x))\cos(\sin x)\cos x.\end{align}$$

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    $\begingroup$ @Sebastiano Done. $\endgroup$ – J.G. Jan 23 at 21:07

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