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I'm a student taking my calculus class. My question is simple but I couldn't find the answer anywhere. The question is how do I differentiate

$$ \alpha = g . x $$

with respect to $x$ if $\alpha$ is a scalar and $g$ is a 1 by n vector and $x$ is a n by 1 vector?

I found differentiation results for many types of equations similar to the above but none of them are in the above form. Please help.

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    What do you mean by taking the derivative of a function $\alpha$ with respect to a vector? Are you talking about the gradient, $\nabla \alpha$? – 2017-02-15
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    (1) Differentiate with respect to what? $\mathbf x$ ($=(x,y,z)^T$?) or $t$? (2) Is $\mathbf g$ a function (of $t$ or $(x,y,z)$) or a constant vector? – 2017-02-15
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    @MichaelSeifert yes I guess it's the gradient. I've found a resource [link](http://www.atmos.washington.edu/~dennis/MatrixCalculus.pdf) if that help to explain what I'm after. topic 5 explains differentiation but doesn't contain this format. – 2017-02-15
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    Are you thinking of the "matrix calculus" sense? See [derivative of scalar-by-vector](https://en.wikipedia.org/wiki/Matrix_calculus). – 2017-02-15
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    @user3658307 yes in matrix calculus. I've found [this](http://www.atmos.washington.edu/~dennis/MatrixCalculus.pdf) resource but it doesn't show the above form (see topic 5 starting from page 4). – 2017-02-15
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    @Bye_World it's respect to x and g is a constant. There's no t here. – 2017-02-15
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    @PPGoodMan Then look at proposition 7 in that document you linked to and take $A=I$, the identity matrix. That is exactly what you're asking for. – 2017-02-15
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    @Bye_World thanks for pointing that out. And sorry for being so dumb not to think of that. I'm just a beginner. – 2017-02-15
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    No problem. :-) – 2017-02-15

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As the comments above note, if: $$ \alpha = u^Tv = u\cdot v $$ where $u,v\in\mathbb{R}^n$ are column vectors, then: $$ \frac{\partial \alpha}{\partial v} = u^T $$ This is because (by definition in matrix calculus) the derivative of a scalar function with respect to a vector is: $$ \frac{\partial \alpha}{\partial v}=\left(\frac{\partial \alpha}{\partial v_1},\frac{\partial \alpha}{\partial v_2},\ldots,\frac{\partial \alpha}{\partial v_n} \right) \in \mathbb{R}^n $$ This can be derived by looking at each component: $$ \frac{\partial \alpha}{\partial v_i} = \frac{\partial}{\partial v_i}\sum\limits_j u_jv_j = \sum\limits_j u_j\frac{\partial v_j}{\partial v_i} =\sum\limits_ju_j\delta_{ij}=u_i $$ where $\delta_{ij}$ is the Kronecker delta.