y(t) = -0.2451 cos 4t + 0.9804 sin 4t + e^(-3t ) (A cos⁡〖t+B〗 sin⁡〖t)〗

y'(t) = ?

You are not being asked to solve a differential equation. You only have to take the derivative of the y(t) function you are given.

Just add the derivatives of the three separate terms.

The derivative of
-0.2451 cos(4t)
is
0.9804 sin(4t)

Your third term is illegible.

To find the derivative of y(t), we need to use the rules of differentiation.

Given that y(t) = -0.2451cos(4t) + 0.9804sin(4t) + e^(-3t)(Acos(t+B)sin(t)), we can differentiate each term separately and then sum up the results.

Let's take it step by step:

1. Differentiating -0.2451cos(4t):
The derivative of cos(x) with respect to x is -sin(x), so the derivative of -0.2451cos(4t) with respect to t is:
-0.2451 * (-sin(4t)) = 0.2451sin(4t).

2. Differentiating 0.9804sin(4t):
The derivative of sin(x) with respect to x is cos(x), so the derivative of 0.9804sin(4t) with respect to t is:
0.9804 * cos(4t) = 0.9804cos(4t).

3. Differentiating e^(-3t)(Acos(t+B)sin(t)):
To differentiate this term, we need to use the product rule. The product rule states that if we have two functions u(x) and v(x), the derivative of their product u(x)v(x) with respect to x is given by:
u'(x)v(x) + u(x)v'(x).

Let's apply the product rule to e^(-3t) and (Acos(t+B)sin(t)):
u(t) = e^(-3t) and v(t) = Acos(t+B)sin(t).
u'(t) = -3e^(-3t) (derivative of e^(-3t) with respect to t)
v'(t) = (-Asin(t+B)sin(t) + Acos(t+B)cos(t)) (product rule applied to Acos(t+B)sin(t))

Using the product rule, the derivative of e^(-3t)(Acos(t+B)sin(t)) with respect to t is:
u'(t)v(t) + u(t)v'(t) = (-3e^(-3t))(Acos(t+B)sin(t)) + (e^(-3t))(-Asin(t+B)sin(t) + Acos(t+B)cos(t))

4. Summing up the results:
Now, let's add up all the derivatives we found:
y'(t) = 0.2451sin(4t) + 0.9804cos(4t) + (-3e^(-3t))(Acos(t+B)sin(t)) + (e^(-3t))(-Asin(t+B)sin(t) + Acos(t+B)cos(t))

So, the derivative of y(t) is 0.2451sin(4t) + 0.9804cos(4t) + (-3e^(-3t))(Acos(t+B)sin(t)) + (e^(-3t))(-Asin(t+B)sin(t) + Acos(t+B)cos(t)).