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$A$ is just a scalar.

Should the derivative of $Ax$ be $A$ or $A^d$?

Context: I'm trying to evaluate this integral

$$\int\varphi(2^n(x-y))\varphi(2^n(x-y))\mathrm{d}x$$

and I know that

$$\int\varphi(2^n(x-y))\varphi(2^n(x-y))\mathrm{d}(2^n(x-y))=1.$$

$\varphi$ is a function from $\mathbb{R}^d\to \mathbb{R}$.

Since

$$\frac{\mathrm{d}}{\mathrm{d}x}2^n(x-y) = 2^n,$$ we have $$2^{-n}\mathrm{d}(2^n(x-y))= \mathrm{d}x.$$

Substituting into the original integral,

$$\int\varphi(2^n(x-y))\varphi(2^n(x-y))\mathrm{d}x$$ $$=\int\varphi(2^n(x-y))\varphi(2^n(x-y))2^{-n}\mathrm{d}(2^n(x-y))$$ $$=2^{-n}.$$

Have I done this substitution correctly?

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    At any point $x$, the derivative is the map $h\in \Bbb R^n \mapsto Ah$, i.e. the map itself. – 2017-01-29
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    @OpenBall Which is true even when $A$ is a matrix. – 2017-01-29
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    @Arthur sure ${}$ – 2017-01-29
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    @OpenBall thanks for your comment. I've edited the question to add some more context, is what I've done alright? – 2017-01-29

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Your problem is not computing the derivative, but applying correctly the change of variables formula.

We have that: $$D Ax = A$$ which can be simply interpreted as: $$\frac{d}{dx_j}(Ax)_{i} = A_{i,j}$$ In the examples you consider we just have $A = \lambda I,$ since we are looking at substitutions of the kind $x = \lambda y.$ Now I invite you to check out the Change of Variable Formula. Then you will see that $$dx = |det(\lambda I)| dy = \lambda^d dy.$$