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I assume your matrix A has entries that do not depend on the entries of x? Then with $y=Ax$ you already define $y$ to be a linear function in $x$, so the derivative of $y$ with respect to $x$ is simply $x\mapsto A$.

Similarly, assuming that $A$ a symmetric matrix that is independent of $x$ (otherwise your function doesn't deserve the name quadratic form) you have that the total derivative of $\alpha$ with respect to $x$ is simply $x\mapsto x^t A$.

No computation necessary; just a calculus book.