how to evaluate derivative of convolution integral?
Note that option (b) is not correct, and that it is also not equal to what you came up with. Option (b) is just the multiplication of $x(t)$ and $y'(t)$, not the convolution. Your solution and option (c) are both correct, assuming that all derivatives exist and that the convolution integrals converge, because with that assumption the following holds:

$$\frac{d}{dt}\left(x(t)\star y(t)\right)=x'(t)\star y(t)=x(t)\star y'(t)\tag{1}$$

where $x'(t)$ and $y'(t)$ denote the derivatives of $x(t)$ and $y(t)$, respectively.

An ideal differentiator is just an LTI system, so it can be written as a convolution, and since convolution is associative, $(1)$ must hold.

This can also most easily be seen in the frequency domain. Differentiation corresponds to multiplication with $j\omega$, so we have

$$j\omega\left(X(\omega)Y(\omega)\right)=\left(j\omega X(\omega)\right)Y(\omega)=X(\omega)\left(j\omega Y(\omega)\right)\tag{2}$$

which is equivalent to $(1)$.