Derivative of convolution of distribution and test function

Question

If $\phi$ is a test function (compactly supported, smooth) on $\mathbb R^d$ and T a distribution, is the following calculation correct? $$(\partial{x_j}\phi)*T(x)=T((\partial{x_j}\phi)(x-.))=\partial{x_j}T(-\phi(x-.))=(-\phi)*\partial{x_j}T$$ where $\phi(x-.)$ denotes the function $y \to \phi(x-y)$ and the convolution of a test function and a distribution is a function in x whose value is the value of the distribution evaluated at the above function. I'm asking since in my notes the equality of the first and last term is said to be without the minus sign, so Im not quite sure whether Im doing smth wrong...