prove $\langle f(x),f'(x)\rangle = 0$

Question

Let $f: R \to R^n$ be a differentiable function such that $\forall x \quad||f(x)|| = 1$

prove that $\forall x \quad \langle f(x),f'(x)\rangle = 0$

i thought of the following proof but not sure it is correct

mark $f(x) = (y_1,y_2 \ldots y_n )$

so using the definition of deritative we get:

$\langle f(x),f'(x)\rangle = \langle f(x), \lim _{h\to0} \frac{f(x+h) - f(x)}{h}\rangle = \lim _{h\to0} \frac{f(x)f(x+h) - f(x)f(x)}{h} = \lim _{h\to0} \frac{(y_1, \ldots ,y_n)(y_1 +o(h), \ldots ,y_n+o(h)) - 1} {h} = \lim_{h\to0} \frac{y_1^2 + \ldots + y_n^2 + o(h)(y_1 +\ldots+ y_n) -1}{h} = \lim_{h\to0} \frac{o(h)(y_1 + \ldots + y_n)}{h} = 0$

is this proof ok? thx

Answer 1

the propsed proof is incorrect

I relied on the fact:

$$ f(x+h) = f(x) + o(h)$$

which is wrong because

$$f(x+h) = f(x) + o(1) = f(x)+f'(x)+o(h)$$

Answer 2

The proposed proof seems correct, but I think this is easier:

Take $\langle f(x),f(x)\rangle = 1$ and differentiate both sides to get directly the result, since the product rule is also valid for the dot product.