The problem asks us to determine the unilateral Laplace transform of the function t f(t), given that the unilateral Laplace transform of f(t) is \(F(s) = \frac{1}{s^2+s+1}\).

Laplace Transform Property for t f(t)

To solve this question, we utilize a fundamental property of the Laplace transform known as the "differentiation in the s-domain" property. This property establishes a relationship between the Laplace transform of a function multiplied by t and the derivative of its Laplace transform with respect to s.

The property states that if \(L\{f(t)\} = F(s)\), then the Laplace transform of t f(t) is given by the following formula:

\[L\{t f(t)\} = -\frac{d}{ds}F(s)\]

In this specific problem, we are provided with the expression for \(F(s)\):

\[F(s) = \frac{1}{s^2+s+1}\]

Differentiating F(s) with Respect to 's'

Our next step is to find the derivative of \(F(s)\) concerning s. To make the differentiation easier, we can rewrite \(F(s)\) using a negative exponent:

\[F(s) = (s^2+s+1)^{-1}\]

Now, we will calculate \(\frac{d}{ds}F(s)\):

\[\frac{d}{ds} \left( \frac{1}{s^2+s+1} \right) = \frac{d}{ds} (s^2+s+1)^{-1}\]

We will apply the chain rule for differentiation. The chain rule states that if you have a function of the form \([g(x)]^n\), its derivative is \(n \cdot [g(x)]^{n-1} \cdot g'(x)\).

In our case:

• The inner function \(g(s) = s^2+s+1\).
• The exponent \(n = -1\).

First, let's find the derivative of the inner function \(g(s)\) with respect to s:

\[g'(s) = \frac{d}{ds}(s^2+s+1)\]

\[g'(s) = 2s+1\]

Now, substitute this into the chain rule formula:

\[\frac{d}{ds} (s^2+s+1)^{-1} = (-1) \cdot (s^2+s+1)^{-1-1} \cdot (2s+1)\]

\[ = (-1) \cdot (s^2+s+1)^{-2} \cdot (2s+1)\]

\[ = -\frac{2s+1}{(s^2+s+1)^2}\]

Final Laplace Transform of t f(t)

We have determined that \(\frac{d}{ds}F(s) = -\frac{2s+1}{(s^2+s+1)^2}\).

Now, we substitute this result back into the main formula for \(L\{t f(t)\}\):

\[L\{t f(t)\} = -\frac{d}{ds}F(s)\]

\[L\{t f(t)\} = - \left( -\frac{2s+1}{(s^2+s+1)^2} \right)\]

When we multiply by \(-1\), the negative sign cancels out:

\[L\{t f(t)\} = \frac{2s+1}{(s^2+s+1)^2}\]

Comparing with Provided Options

Let's review the given options and compare them with our calculated Laplace transform:

• Option 1: \(-\frac{s}{(s^2+s+1)^2}\)
• Option 2: \(-\frac{2s+1}{(s^2+s+1)^2}\)
• Option 3: \(\frac{s}{(s^2+s+1)^2}\)
• Option 4: \(\frac{2s+1}{(s^2+s+1)^2}\)

Our calculated Laplace transform for t f(t), which is \(\frac{2s+1}{(s^2+s+1)^2}\), perfectly matches Option 4.