Please help me solve this question:
let $f:R^n \rightarrow R$ be a function. for all differentiable curve $\gamma: [-1,1] \rightarrow R^n$ such that $\gamma(0) = \vec{0} \space$, $f \circ \gamma :[-1,1] \rightarrow R^n $ is differentiable in $t=0$ . Prove that $f$ is differentiable in $\vec{0}$.
My way of thought here is to use the chain rule to say that that $(f \circ \gamma)'(0) =\nabla f(\gamma(0)) \cdot \gamma'(0)$ and than somehow to show that: $ \lim_{\vec{x} \to \vec{0} }\frac{f(x_1,...,x_n)-f(\vec{0})- \nabla f(\vec{0}) \cdot (x_1,...,x_n)} {||(x_1,...,x_n)||}$
I'm not really sure how to proceed from here' so it will be grat if you could have helped me.
thanks.