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Let $f_j:\mathbb{R}\rightarrow \mathbb{R}$ for $j=1,2,\ldots,n$ be continuously differentiable functions. Prove that the function $f(x_1,x_2,\ldots,x_n) = f_1(x_1)\cdots f_n(x_n)$ is differentiable on $\mathbb{R}^n$

How do we prove this since each $f_j$ is differential then composition of this functions again differentiable and I don't where I should start this question. Can any one help me please?

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    Try applying the product rule for derivatives. – 2017-02-13
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    It's customary to use three dots (an "ellipsis") when omitting; so common, in fact, that the command `\ldots` will produce three appropriately spaced dots (where `l`, for **l**ower, may be replaced with `c` for a horizontally **c**entered version). – 2017-02-13

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We have that $$f=\prod_{i=1}^n f_i \circ \pi_i,$$ where $\pi_i$ are the $i$-projections. Since composition of differentiable functions is differentiable, each $f_i \circ \pi_i$ is differentiable. Since products of differentiable functions are differentiable, $f$ also is.

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    Doesn't it need a contintuity argument: since they are "continuisly" differentiable.... – 2017-02-13
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Generally speaking, the product of continuously differentiable functions will also be differentiable. For example, let $\quad$$f(x) = x^2$,$\quad$ $f(y)= y^3$,$\quad$ $f(z) = z^4$

Then:

$f(x,y,z) = x^2y^3z^4$

Which is certainly differentiable in $\mathbb{R}^3$.

Since, $\nabla$$f$(x,y,z) = $\begin{bmatrix}2xy^3z^4 & 3y^2x^2z^4 & 3z^3x^2y^3\end{bmatrix}$

Try to generalize this example to $\mathbb{R}^n$.