Without assumptions on the existence of a derivative for $y(t)$, you can easily rule out (b) and (d), because they have no sense. With some notions of unit homogeneity, you can rule (d) out again. If you take a wild $y(t)$, like the Dirac function, the product in (a) is not defined. While the convolution is OK, since the Dirac function is the neutral element for convolution.

Thus, the only likely solution is (c), for which, under suitable conditions, Matt's answer is valid.

The property illustrated here is that the derivation is distributive over the convolution, hence, if $f$ is differentiable:

$$ \frac{d}{dt}(f(t)*g(t)) = (\frac{d}{dt}f(t))*g(t)$$

You can check for instance:

• Derivative of convolution

Without assumptions on the existence of a derivative for $y(t)$, you can easily rule out (b) and (d), because they have no sense. With some notions of unit homogeneity, you can rule (d) out again. If you take a wild $y(t)$, like the Dirac function, the product in (a) is not defined. While the convolution is OK, since the Dirac function is the neutral element for convolution.

Thus, the only likely solution is (c), for which, under suitable conditions, Matt's answer is valid.

The property illustrated here is that the derivation is distributive over the convolution, hence, if $f$ is differentiable:

$$ \frac{d}{dt}(x(t)*y(t)) = (\frac{d}{dt}x(t))*y(t)$$

You can check for instance:

• Derivative of convolution