Let X(ω) be the Fourier transform of the signal

$x(t)=e^{-t^4}\cos t,-\mathrm{\infty }<t<\mathrm{\infty }.$

The value of the derivative of X(ω) at ω = 0 is __________ (Rounded off to 1 decimal place).

The correct answer is 0.

To understand why the derivative of the Fourier transform $X\left(\omega \right)$ evaluated at $\omega =0$ is equal to 0, we can analyze the symmetry properties of the signal $x\left(t\right)$.

First, let's write down the given signal:
$x\left(t\right)=e^{-t^4}\cos \left(t\right)$
for all real values of $t$ (i.e., $-\infty <t<\infty$).

We check if the signal $x\left(t\right)$ is even or odd by substituting $-t$ for $t$:
$x(-t)=e^{-{(-t)}^4}\cos (-t)$

Since ${(-t)}^4=t^4$ and $\cos (-t)=\cos \left(t\right)$, we have:
$x(-t)=e^{-t^4}\cos \left(t\right)=x\left(t\right)$

This shows that the signal $x\left(t\right)$ is an even function of time.

Next, let's look at the relationship between a signal $x\left(t\right)$ and the derivative of its Fourier transform $X\left(\omega \right)$. The Fourier transform is defined as:
$X\left(\omega \right)={\int }_{-\infty }^{\infty }x\left(t\right)e^{-j\omega t}dt$

Differentiating both sides with respect to $\omega$ under the integral sign gives:
$\frac{dX\left(\omega \right)}{d\omega }={\int }_{-\infty }^{\infty }x\left(t\right)(-jt)e^{-j\mathrm{\omega t}}dt$

Evaluating this derivative at $\omega =0$ yields:
${\left.\frac{dX\left(\omega \right)}{d\omega }\right|}_{\omega =0}=-j{\int }_{-\infty }^{\infty }tx\left(t\right)dt$

Let's define a new integrand function: $g\left(t\right)=tx\left(t\right)$. We determine the symmetry of $g\left(t\right)$:
$g(-t)=(-t)x(-t)$

Since $x\left(t\right)$ is an even function, we have $x(-t)=x\left(t\right)$, which gives:
$g(-t)=-tx\left(t\right)=-g\left(t\right)$

Thus, $tx\left(t\right)$ is an odd function of time.

An important mathematical property of odd functions integrated over a symmetric interval around zero is that the integral is zero:
${\int }_{-\infty }^{\infty }tx\left(t\right)dt=0$

Substituting this result back into our derivative expression:
${\left.\frac{dX\left(\omega \right)}{d\omega }\right|}_{\omega =0}=-j\left(0\right)=0$

Therefore, the value of the derivative of the Fourier transform $X\left(\omega \right)$ at $\omega =0$ is exactly 0.