Suppose that f\!\left(\frac{\pi}{4}\right)=-4 and f'\!\left(\frac{\pi}{4}\right)=7, and let g(x) = f(x) \sin x and \displaystyle h(x) = \frac{\cos x}{f(x)}. Answer the following questions.

1. Find g'(\pi/4).
Answer: g'(\pi/4) = 
2. Find h'(\pi/4).
Answer: h'(\pi/4) =

Let's solve both questions step by step.

### 1. Finding \( g'(\pi/4) \)

We are given:
- \( g(x) = f(x) \sin x \),
- \( f\left(\frac{\pi}{4}\right) = -4 \),
- \( f'\left(\frac{\pi}{4}\right) = 7 \).

To find \( g'(x) \), we use the product rule:
\[
g'(x) = f'(x) \sin x + f(x) \cos x
\]
Now, substitute \( x = \frac{\pi}{4} \):
\[
g'\left(\frac{\pi}{4}\right) = f'\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right) + f\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{4}\right)
\]
We know that:
\[
\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
\]
Substituting the values:
\[
g'\left(\frac{\pi}{4}\right) = 7 \cdot \frac{\sqrt{2}}{2} + (-4) \cdot \frac{\sqrt{2}}{2}
\]
Simplifying:
\[
g'\left(\frac{\pi}{4}\right) = \frac{7\sqrt{2}}{2} - \frac{4\sqrt{2}}{2}
\]
\[
g'\left(\frac{\pi}{4}\right) = \frac{3\sqrt{2}}{2}
\]

### 2. Finding \( h'(\pi/4) \)

We are given:
- \( h(x) = \frac{\cos x}{f(x)} \).

To find \( h'(x) \), we use the quotient rule:
\[
h'(x) = \frac{f(x)(-\sin x) - \cos x f'(x)}{(f(x))^2}
\]
Now, substitute \( x = \frac{\pi}{4} \):
\[
h'\left(\frac{\pi}{4}\right) = \frac{f\left(\frac{\pi}{4}\right)(-\sin\left(\frac{\pi}{4}\right)) - \cos\left(\frac{\pi}{4}\right) f'\left(\frac{\pi}{4}\right)}{\left(f\left(\frac{\pi}{4}\right)\right)^2}
\]
Substituting the known values:
\[
h'\left(\frac{\pi}{4}\right) = \frac{-4 \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \cdot 7}{(-4)^2}
\]
\[
h'\left(\frac{\pi}{4}\right) = \frac{-2\sqrt{2} - \frac{7\sqrt{2}}{2}}{16}
\]
Combine the terms in the numerator:
\[
h'\left(\frac{\pi}{4}\right) = \frac{-\frac{4\sqrt{2}}{2} - \frac{7\sqrt{2}}{2}}{16} = \frac{-\frac{11\sqrt{2}}{2}}{16}
\]
Simplify:
\[
h'\left(\frac{\pi}{4}\right) = \frac{-11\sqrt{2}}{32}
\]

### Final Answers:
1. \( g'\left(\frac{\pi}{4}\right) = \frac{3\sqrt{2}}{2} \)
2. \( h'\left(\frac{\pi}{4}\right) = \frac{-11\sqrt{2}}{32} \)

Would you like more details or clarifications on these solutions?

Here are 5 related questions to expand on this:
1. How is the product rule applied to find derivatives of products of functions?
2. What is the significance of the quotient rule in calculus?
3. Can we interpret the physical meaning of \( g'(x) \) and \( h'(x) \) in certain contexts?
4. How would the derivatives change if \( f\left(\frac{\pi}{4}\right) \) or \( f'\left(\frac{\pi}{4}\right) \) had different values?
5. Can you apply implicit differentiation to find the derivatives of more complex functions?

**Tip:** Always check whether you are dealing with a product or quotient of functions before choosing the correct differentiation rule.

Suppose that f(π/4) = -4 and f'(π/4) = 7, and let g(x) = f(x)sin(x) and h(x) = cos(x)/f(x). Answer the following questions.

1. Find g'(π/4).
2. Find h'(π/4).

Learn how to find the derivatives of g(x) = f(x)sin(x) and h(x) = cos(x)/f(x) using the product and quotient rules. Detailed step-by-step solutions for finding g'(π/4) and h'(π/4), involving trigonometric identities and differentiation rules. Suitable for advanced high school calculus students.

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