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 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:

$$x'(t)\star y(t)=x(t)\star y'(t)\tag{1}$$

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 $(1)$.

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 $(1)$.

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 its 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:

$$x'(t)\star y(t)=x(t)\star y'(t)\tag{1}$$

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 $(1)$.

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 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:

$$x'(t)\star y(t)=x(t)\star y'(t)\tag{1}$$

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 $(1)$.