To find the derivative of the function $f(t) = \frac{t\sin(t)}{2t + 1}$, we can use the quotient rule. The quotient rule states that if you have a function $f(t) = \frac{g(t)}{h(t)}$, then the derivative $f'(t)$ is given by:

$f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{[h(t)]^2}$

Here, $g(t) = t\sin(t)$ and $h(t) = 2t + 1$. Let's find their derivatives:

1. Derivative of $g(t) = t\sin(t)$: $g'(t) = \frac{d}{dt} [t\sin(t)] = t\cos(t) + \sin(t)$ (using the product rule).

2. Derivative of $h(t) = 2t + 1$: $h'(t) = \frac{d}{dt} [2t + 1] = 2$

Now, apply the quotient rule:

$f'(t) = \frac{(t\cos(t) + \sin(t))(2t + 1) - t\sin(t)(2)}{(2t + 1)^2}$

Simplifying the numerator:

$f'(t) = \frac{2t^2\cos(t) + t\cos(t) + 2t\sin(t) + \sin(t) - 2t\sin(t)}{(2t + 1)^2}$

Further simplification gives:

$f'(t) = \frac{2t^2\cos(t) + t\cos(t) + \sin(t)}{(2t + 1)^2}$

So, the derivative of the function $f(t) = \frac{t\sin(t)}{2t + 1}$ is:

$f'(t) = \frac{2t^2\cos(t) + t\cos(t) + \sin(t)}{(2t + 1)^2}$

Would you like more details or have any questions about this process?

Related Questions:

1. How do you apply the product rule in differentiation?
2. What is the physical meaning of the derivative in the context of motion?
3. Can you show the steps for differentiating $g(t) = t\sin(t)$ in detail?
4. How does the quotient rule differ from the product rule?
5. How do you determine when to use the quotient rule vs. the chain rule?

Tip:

Always simplify the derivative step-by-step to avoid errors, especially when using rules like the quotient rule.