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The question is here

Let $f(x_1, \dots , x_n)$ be a function of class $C^1$ on the ball $(x_1)^2 + \cdots + (x_n)^2 < 1$, such that all the first order all partial derivatives are zero at every point of the ball. Prove that $f$ is constant.

My instructor gives a hint that '$(x_1,...,x_{j−1},t\cdot x_j,x_{j+1},...,x_n)$ is in the ball, so for $0 ≤ t ≤ 1$, the derivative of the function $g(t) = f(x_1,...,x_{j−1},t\cdot x_j,x_{j+1},...,x_n)$ with respect to $t$ is zero.' However, this doesn't make any sense to me. Anyone can give me more hints or ideas?

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    What is the derivative of $f(x_1, x_2, \ldots x_{j-1}, tx_j, x_{j+1}, \ldots x_n)$ with respect to $t$? It uses chain rule. – 2017-02-19
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    It should be (x_j)* the partial derivative of f(x_1,...,x_j-1,t*x_j,x_j+1,...,x_n) right? – 2017-02-19
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    Correct, which means that it is 0, whatever $t$ we choose. Now $g(t)$ is a $C^1$ function such that $g'(t)=0$ for all $g$. What does that mean $g$ must be? – 2017-02-19
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    g(t) should be a constant function, but then? I don't understand what we can get from the conclusion that g is constant and why we need to use t*x_j – 2017-02-19
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    Perhaps it's better to use $f(tx_1, tx_2, \ldots tx_n)$ instead. – 2017-02-19
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    To be honest I don't understand why limit the result to the radius $1$ ball or the significance of the hint. I'd prove it by induction on $n$. For example, for $n=2$ and $f\colon \mathbb R^2\to \mathbb R$, if $\partial _1f$ is zero, then $\forall (x,y)\in \mathbb R^2(f(x,y)=\phi(y))$ for some differentiable function $\phi\colon \mathbb R\to \mathbb R$. Then, because $\partial _2$ is null, then $\phi'$ is null which implies that $\phi$ is constant and therefore $f$ is constant. – 2017-02-19

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