Let's solve the derivatives of both functions $f(x)$ and $g(t)$.

(a) For $f(x) = \frac{x - \sqrt{x}}{x^2}$

To find $f'(x)$, we will use the quotient rule for derivatives:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$ where $u(x) = x - \sqrt{x}$ and $v(x) = x^2$.

First, we calculate $u'(x)$ and $v'(x)$:

• $u(x) = x - \sqrt{x} = x - x^{1/2}$ $u'(x) = 1 - \frac{1}{2}x^{-1/2}$
• $v(x) = x^2$ $v'(x) = 2x$

Now, applying the quotient rule: $f'(x) = \frac{(1 - \frac{1}{2}x^{-1/2}) \cdot x^2 - (x - \sqrt{x}) \cdot 2x}{(x^2)^2}$

Simplifying this expression will give the final derivative.

(b) For $g(t) = \frac{\tan(t)}{1 + t}$

We again use the quotient rule where $u(t) = \tan(t)$ and $v(t) = 1 + t$.

• $u(t) = \tan(t)$ $u'(t) = \sec^2(t)$
• $v(t) = 1 + t$ $v'(t) = 1$

Now, applying the quotient rule: $g'(t) = \frac{\sec^2(t) \cdot (1 + t) - \tan(t) \cdot 1}{(1 + t)^2}$

Simplifying this expression will give the final derivative.

Would you like the detailed simplifications for these derivatives? Let me know!

Here are five follow-up questions related to these derivatives:

1. What happens when the quotient rule is used in higher-order derivatives?
2. How does the chain rule interact with the quotient rule when nested functions are involved?
3. How can implicit differentiation be used in functions with multiple variables?
4. How would we approach derivatives of parametric equations?
5. How can L'Hôpital's rule help in evaluating limits that result in indeterminate forms involving derivatives?

Tip: Always double-check the domain of the function after taking derivatives, as differentiating might introduce restrictions (e.g., division by zero).