I need to express the derivative of $f^{(n)}(x)$ in terms of $f'(x)$, meaning $f$ composed with itself n times.
I was able to express $f(f(x))=f'(f(x))f'(x)$
Is the derivative of the composition $\prod_{k=0}^{n-1} f'(f^{(k)}(x))$ ?
Please help
I need to express the derivative of $f^{(n)}(x)$ in terms of $f'(x)$, meaning $f$ composed with itself n times.
I was able to express $f(f(x))=f'(f(x))f'(x)$
Is the derivative of the composition $\prod_{k=0}^{n-1} f'(f^{(k)}(x))$ ?
Please help
Note: Sometimes $f^{(n)}$ means the $n$th derivative. In this context it means the $n$th iterate.
Put $g(x) = f^{(n)}(x)$ in $\frac{d}{dx} f(g(x)) = g'(x) f'(g(x))$ to get $\frac{d}{dx} f^{(n+1)}(x) = [\frac{d}{dx} f^{(n)}(x)] f'(f^{(n)}(x)).$
Using $f^{(0)}(x) = x$ we find, by induction, that $\frac{d}{dx} f^{(n)}(x) = f'(x) \cdot f'(f(x)) \cdot f'(f^{(2)}(x)) \cdots f'(f^{(n-1)}(x)).$