Find the derivatives of the following function and calculate $f'(1)$.

Question

$f(x)=\prod_{n=1}^{100}(tan\frac {\pi x^n}{4}-n)$

I can solve the second question,use $g(x)=\prod_{n=2}^{100}(tan\frac {\pi x^n}{4}-n)$ and I can get $f(x)=(tan\frac {\pi x^n}{4}-n)g(x),n=1$.

But I want to know if there is any other way to solve it?

And how I can solve the first question?

thanks very much.

Answer 1

Since you have \begin{align} f(x) = \prod^{100}_{n=1}\left(\tan\frac{\pi x^n}{4}-n \right) \end{align} then it follows \begin{align} f'(x)=&\ \sum^{100}_{j=1}\left\{ \left(\frac{j\pi x^{j-1}}{4}\sec^2\frac{\pi x^j}{4} \right)\prod^{100}_{n=1\\ n\neq j}\left(\tan\frac{\pi x^n}{4}-n \right)\right\}. \end{align} Setting $x=1$ yields \begin{align} f'(1) =&\ \sum^{100}_{j=1}\left\{ \left(\frac{j\pi }{4}\sec^2\frac{\pi}{4} \right)\prod^{100}_{n=1\\ n\neq j}\left(\tan\frac{\pi}{4}-n \right)\right\}=\sum^{100}_{j=1}\left\{ \frac{j\pi}{2}\prod^{100}_{n=1\\n\neq j}(1-n)\right\}\\ =&\ \frac{\pi}{2}\left\{ \prod^{100}_{n=2}(1-n)\right\} = -\frac{\pi}{2} 99! \end{align}