Answer:
[tex]f'(x)=\displaystyle -\frac{e^x}{(1+e^x)^2}\cos\biggr(\frac{1}{1+e^x}\biggr)[/tex]
Step-by-step explanation:
Recall the power series [tex]\sin(x)=\displaystyle \sum\limits^\infty_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex].
In this case, [tex]x[/tex] is replaced with [tex]\displaystyle \frac{1}{1+e^x}[/tex], so our power series actually works out to be [tex]\displaystyle \sin\biggr(\frac{1}{1+e^x}\biggr) =\sum\limits^\infty_{n=0}(-1)^n\frac{\bigr(\frac{1}{1+e^x}\bigr)^{2n+1}}{(2n+1)!}[/tex]! Amazing, huh?
Now, we find the derivative of the function by using the chain rule:
[tex]\displaystyle \frac{d}{dx}\sin\biggr(\frac{1}{1+e^x}\biggr)=\frac{d}{dx}\biggr(\frac{1}{1+e^x}\biggr)*\cos\biggr(\frac{1}{1+e^x}\biggr)=-\frac{e^x}{(1+e^x)^2}\cos\biggr(\frac{1}{1+e^x}\biggr)[/tex]
You didn't specify if you just wanted the derivative of the series to be a function or a series, so I'm going to assume you want the function. Let me know if there's more to your problem.
