If $f(x) = \sec(x^{333})$ then find $f^{666}(x)$ Where $f^n(x)$ represents the $n$-th derivative of a function.

**Solution**

At first this looks crazy but we quickly notice that a lot of terms in the 666-th derivative will evaluate to zero at $x=0$. More precisely, the first derivative contains factors $\sec(x^{333}), \tan(x^{333}),$ and $x^{332}$.

The tangent and the $x^n$ factors evaluate to zero, but the secant factor evaluates to 1. So all we need to figure out is, at each derivative step, which terms will affect the final derivatives.

To make it clearer, let's generalize into: $f(x) = \sec(x^n)$. $$f^1(x) = nx^{n-1} \sec(x^n) \tan(x^n)$$ $$f^2((x) = n^2x^{2n-2} \sec(x^n) \tan^2(x^n) + n(n-1)x^{n-2} \sec(x^n) \tan(x^n) + n^2 x^{2n-2} \sec^3(x^n)$$ The first term will continue to have higher derivatives with higher powers of $x$ and $\tan(x^n)$. Ditto for the second term. What we need to worry about is only the third term.

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