To solve for $p'(t)$ using the quotient rule, we start with the function:

$p(t) = \frac{\sqrt{t}}{2t - 1}$

Step 1: Rewrite $\sqrt{t}$

Rewrite $\sqrt{t}$ as $t^{1/2}$.

So, $p(t) = \frac{t^{1/2}}{2t - 1}$

Step 2: Apply the Quotient Rule

The quotient rule for derivatives states that if $p(t) = \frac{f(t)}{g(t)}$, then: $p'(t) = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}$

In our case:

• $f(t) = t^{1/2}$ and $g(t) = 2t - 1$.

Derivatives of $f(t)$ and $g(t)$:

1. Find $f'(t)$: $f(t) = t^{1/2} \implies f'(t) = \frac{1}{2}t^{-\frac{1}{2}} = \frac{1}{2\sqrt{t}}$

2. Find $g'(t)$: $g(t) = 2t - 1 \implies g'(t) = 2$

Step 3: Substitute into the Quotient Rule Formula

$p'(t) = \frac{\left(\frac{1}{2\sqrt{t}}\right)(2t - 1) - (t^{1/2})(2)}{(2t - 1)^2}$

Step 4: Simplify the Expression

Expand and simplify the numerator:

$p'(t) = \frac{\frac{2t - 1}{2\sqrt{t}} - 2t^{1/2}}{(2t - 1)^2}$

Convert $2t^{1/2}$ to $\frac{2\sqrt{t}}{1}$ for further simplification if needed.

This question falls under the "calculus" category, as it involves differentiation techniques.

Would you like me to continue with further simplification or explain any part in detail?

Follow-up Questions:

1. How does the quotient rule apply to more complex functions?
2. Can this problem be solved using alternative methods like logarithmic differentiation?
3. How would the derivative change if we added a constant to $p(t)$?
4. What are common mistakes made when applying the quotient rule?
5. How does this compare to using the product rule in differentiation?

Tip:

When using the quotient rule, ensure each derivative (numerator and denominator) is calculated accurately before combining them to avoid errors.