If $n<0$ assume $f(x) \not = 0$. Prove that $(f^n)^\prime(x)=n(f(x))^{n-1}$.

Question

Question: Let $f: I \rightarrow R$ be differentiable. Given $n \in Z$, define $f^n$ be the function defined by $f^n(x)= (f(x))^n$. If $n<0$ assume $f(x) \not = 0$. Prove that $(f^n)^\prime(x)=n(f(x))^{n-1}$.

Let $g: S \rightarrow R$ defined by $g(y) = y^n$ for $n<0$. if I can show that $g$ is differentiable, then I can prove $(f^n)^\prime(x)=n(f(x))^{n-1}$ by using the chain rule.

I saw one solution to prove $g$ is differentiable, using binomial theorem, and I am wondering if there is another way to prove this without using binomial theorem since I haven't seen this in my course.

Thank you in advance.

Answer 1

To show that $f(x) = x^n$ is differentiable, you just need to know that products of differentiable functions are differentiable. This follows from the product rule. For a quick proof:

\begin{align} \lim_{h \to 0}{\frac{f(x+h)g(x+h) - f(x)g(x)}{h}} & = \lim_{h\to 0}{\frac{f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)}{h}} \\ & = \lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}g(x+h)}+\lim_{h\to 0}{\frac{g(x+h)-g(x)}{h}g(x)} \\ & = f'(x)g(x)+g'(x)f(x) \end{align}

Then since $h(x)=x$ is differentiable, by induction so is $g(x)=x^n$ as the $n$-fold product of $h$.

Then apply Chain Rule to the composition $g \circ f$ to get $$(g\circ f)' = f' \cdot (g'\circ f)$$ or in other words $$(f^n)'(x) = f'(x)\cdot n f(x)^{n-1}$$

Answer 2

HINT

You are composing $a(x) = 1/x$ and $b(x) = x^n$.

Answer 3

$$ g:x \mapsto f^n(x)$$ is the composition of $$ f $$ by $$ x \mapsto x^n $$

You should know the derivative of a composition of two functions?

Other method: by induction.

Edit: didn't see you wanted to prove the expression of the derivative of $$ x \mapsto x^n $$ I suggest you use induction then.

Answer 4

$$\lim_{h\rightarrow 0}\frac{f^n(a+h)-f^n(a)}{h}= \lim_{h\rightarrow 0}\frac{(f(a+h)-f(a) ) \sum_{i=0}^{n-1}f^{i}(a+h)f^{n-i}(a)}{h} \\ = \lim_{h\rightarrow 0}\frac{f^n(a+h)-f^n(a)}{h}\cdot\lim_{h\rightarrow0}\sum_{i=0}^{n-1}f^{i}(a+h)f^{n-i}(a) \\ =f'(a)\cdot n f^{n-1}(a)$$

Here we use the identity $$a^n-b^n=(a-b)(a^{n-1}+ba^{n-2}+b^2a^{n-3}\cdots+b^{n-1})$$