Let's go through the problems visible in your image step-by-step.

Part 1: Find the derivatives of each function

You need to find the derivatives of the given functions, without using the limit definition of derivatives. I will walk through each one:

(a) $f(x) = \frac{-\cos(x)}{x}$

This is a quotient rule problem. The quotient rule states that for a function of the form $\frac{u(x)}{v(x)}$, the derivative is: $\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) u'(x) - u(x) v'(x)}{v(x)^2}$ Here:

• $u(x) = -\cos(x)$, so $u'(x) = \sin(x)$
• $v(x) = x$, so $v'(x) = 1$

Applying the quotient rule: $f'(x) = \frac{x \sin(x) - (-\cos(x))}{x^2} = \frac{x \sin(x) + \cos(x)}{x^2}$

(b) $g(x) = \frac{-2e^x}{6x^2} - \csc(x)$

For this function, we break it down:

• First term: $\frac{-2e^x}{6x^2}$ requires the quotient rule:

  • $u(x) = -2e^x$, so $u'(x) = -2e^x$
  • $v(x) = 6x^2$, so $v'(x) = 12x$ Applying the quotient rule here, we get: $\frac{6x^2(-2e^x) - (-2e^x)(12x)}{(6x^2)^2}$ Simplifying: $\frac{-12x^2 e^x + 24x e^x}{36x^4} = \frac{-e^x (12x^2 - 24x)}{36x^4}$

• Second term: Derivative of $-\csc(x)$ is $\csc(x) \cot(x)$.

Combining both, we get: $g'(x) = \frac{-e^x (12x^2 - 24x)}{36x^4} + \csc(x) \cot(x)$

(c) $h(x) = \sin^2(x) + \cos^2(x)$

We simplify the expression before differentiating, noting that $\sin^2(x) + \cos^2(x) = 1$, which is a constant. The derivative of a constant is 0: $h'(x) = 0$

(d) $k(x) = \frac{1 + \tan(x)}{\tan(x)}$

We simplify this before differentiating: $k(x) = \frac{1}{\tan(x)} + 1 = \cot(x) + 1$ Now, take the derivative:

• Derivative of $\cot(x)$ is $-\csc^2(x)$
• Derivative of 1 is 0

Thus: $k'(x) = -\csc^2(x)$

Part 2: Compute $h'(x)$ for the given compositions

In this section, you're given a table of values for functions $f(x)$ and $g(x)$, along with their derivatives, and asked to compute $h'(x)$ for various compositions of $f$ and $g$. Here's how we compute them using the chain rule:

• (a) $h(x) = f(g(x)), a = 0$: Using the chain rule: $h'(x) = f'(g(x)) \cdot g'(x)$ At $x = 0$, $g(0) = 1$, so we look up $f'(1)$ and $g'(0)$ from the table: $h'(0) = f'(1) \cdot g'(0) = 3 \cdot 0 = 0$

• (b) $h(x) = g(f(x)), a = 2$: Chain rule again: $h'(x) = g'(f(x)) \cdot f'(x)$ At $x = 2$, $f(2) = 4$, so we look up $g'(4)$ and $f'(2)$: $h'(2) = g'(4) \cdot f'(2) = -1 \cdot 5 = -5$

Let me know if you'd like me to continue solving parts (c) to (g) or if you need any clarification so far!

Related Questions:

1. What is the chain rule, and why is it crucial for computing the derivative of composite functions?
2. Can you explain the quotient rule in more depth?
3. What happens if the inner function in a composition has a discontinuity?
4. How do trigonometric identities help simplify derivative problems like in part (c)?
5. What are some real-world applications of higher-order derivatives?

Tip: When using the chain rule, always differentiate the outer function first and then multiply by the derivative of the inner function.