Here is my problem: "Suppose f is a differentiable function whose domain is $(-\infty,\infty)$. We define an infinite sequence of functions $f_n(x)$ as follows:
$f_1(x)=f(x), f_2(x)=f(f_1(x))$, and so on. That is, $f_n(x)= f(f_{n-1}(x))$ for $n\geq 2$.
State an explicit formula for $\frac{d}{dx}[f_n(x)]$ in which the only derivative is $f'$ and then prove that your formula is correct using Mathematical Induction"
So far, I have found that $\frac{d}{dx} f_2(x)= f'(f(x))\cdot f'(x)$
$\frac{d}{dx} f_3(x)= f'(f(f(x)))\cdot f'(f(x))\cdot f'(x)$
I saw a pattern an my formula is $f'(x)\cdot f'(f(x))\cdot f'(f(f(x)))\cdots f'(f(\dots(f(x))\dots))$ for as large n is.
I am not sure how to prove this using induction though.... Thanks
Heh. You're un UMTYMP, right? :)
Here's a hint:
Try to find $~\frac{d}{dx}[f_2(x)]~$ first, then $~\frac{d}{dx}[f_3(x)]~$. You should see a pattern that is provable by induction.
Edit: Write the formula using big pi notation. Assume this formula works for n=k. Now, prove that it still holds true when you plug in $~n=k+1~$ using the previous assumption.