How do I take the derivative of $\int g(x - y) \mathrm dx$ with respect to $y$?

Question

How do I take the derivative of $\displaystyle\int g(x - y)\,\mathrm dx$ with respect to $y$? Is it just $-g(x - y)\,\mathrm dx$?

Thanks.

Answer 1

Let $$G(x)=\int g(x)dx$$therefore $$G(x-y)=\int g(x-y)dx$$so we have $$\dfrac{d}{dy}\int g(x-y)dx=-g(x-y)$$

Answer 2

Yes. Consider $x$ a constant parameter.

Answer 3

Differentiating under the integral sign, $$\frac{d}{dy}\int g(x-y)dx=\int\partial_y g(x-y)dx=-\int g'(x-y)dx=C(y)-g(x-y).$$Note that since the indefinite integral $\int g(x-y)dx$ is determined up to an integration constant that can depend on $y$, the integration constant $C$ is also a general function of $y$.