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If the signal $x(t)$ has ordinary first derivative $\dot x(t)$, then $\dfrac{d}{dt}\big(x(t)\star y(t)\big)$ is:

  • (a) $\dot x(t)y(t)$
  • (b) $x(t)\dot y(t)$
  • (c) $\dot x(t)\star y(t)$
  • (d) $\dot x(t)\star \dot y(t)$

I solved it as follows: we know,

$$I(t)=x(t)\star y(t)=\displaystyle\int_{-\infty}^{\infty}x(\tau).y(t-\tau) d{\tau}\implies$$ $$\begin{align}\dfrac{dI(t)}{dt}&=\dfrac{d}{dt}\displaystyle\int_{-\infty}^{\infty}x(\tau)y(t-\tau) d{\tau}\\&=\displaystyle\int_{-\infty}^{\infty}\dfrac{\partial}{\partial t}x(\tau).y(t-\tau) d{\tau} \\&=\displaystyle\int_{-\infty}^{\infty}x(\tau).\dfrac{\partial}{\partial t}y(t-\tau) d{\tau}=x(t)\star \dot y(t)\end{align}$$

So option $(b)$ should be correct but answer in the book is given as $(c)$ but I think both options $(b)$ and $(c)$ should be correct because in above method I fixed $x(t)$ and shifted $y(t)$ but we can also do that other way round fixing $y(t)$ and shifting $x(t)$,and we know by doing so,convolution integral remain unaffected,so tell where i'm wrong ,any help would be appreciated

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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 $(1)$.

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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:

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    $\begingroup$ @ Laurent Duval .....thank you sir,.....But if i assume y(t) also as continiuous and smooth function which is infinitely differentiable ....then,i suppose, (d) can't be ruled out $\endgroup$ – Faraday Pathak Jan 28 '18 at 20:14
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    $\begingroup$ It can be then with the 2nd argument. $\endgroup$ – Laurent Duval Jan 28 '18 at 20:22
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    $\begingroup$ Why is (a) not defined if $y(t)=\delta(t)$? $\endgroup$ – Matt L. Jan 28 '18 at 20:51
  • $\begingroup$ Which of course raises the question as to why it is not $y(t)$ that gets differentiated when the differentiation distributes over the convolution. $\endgroup$ – Dilip Sarwate Jan 28 '18 at 22:49

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