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Let’s break down the questions and solve them step by step.

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**3.95**  
Prove the product rule:  
\[
(k \cdot u(x))' = k \cdot u'(x)
\]  

### Solution:
The derivative of a constant \(k\) multiplied by a function \(u(x)\) is obtained using the linearity of derivatives:
\[
\frac{d}{dx}[k \cdot u(x)] = k \cdot \frac{d}{dx}[u(x)] = k \cdot u'(x).
\]
Thus, the result is proved.

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**3.96 (a)**  
We are given \(f(x) = u \cdot v\), where \(u\) and \(v\) are functions of \(x\), and are asked to derive:
\[
\frac{d}{dx} [\ln f(x)] = \frac{u'v + uv'}{uv}.
\]

### Solution:
1. Apply the chain rule to \(\ln f(x)\):
   \[
   \frac{d}{dx} [\ln f(x)] = \frac{1}{f(x)} \cdot f'(x).
   \]
2. Substituting \(f(x) = u \cdot v\), we need \(f'(x)\). Using the product rule:
   \[
   f'(x) = u'v + uv'.
   \]
3. Substituting \(f(x)\) and \(f'(x)\):
   \[
   \frac{d}{dx} [\ln f(x)] = \frac{1}{uv} \cdot (u'v + uv') = \frac{u'v + uv'}{uv}.
   \]
Thus, the result is verified.

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**3.96 (b)**  
Using the result from (a), derive the product rule for \(f(x) = u \cdot v\).  

### Solution:
From (a), we have:
\[
\frac{d}{dx} [\ln f(x)] = \frac{u'v + uv'}{uv}.
\]
Multiplying through by \(uv\) (the denominator of the right-hand side):
\[
f'(x) = u'v + uv'.
\]
Thus, the product rule is derived.

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**3.96 (c)**  
Use a similar approach to derive the quotient rule for \(\frac{u}{v}\).  

### Solution:
For \(f(x) = \frac{u}{v}\), we apply the same logarithmic differentiation method:
\[
\ln f(x) = \ln u - \ln v.
\]
Differentiating both sides:
\[
\frac{d}{dx} [\ln f(x)] = \frac{d}{dx} [\ln u - \ln v].
\]
Using the chain rule:
\[
\frac{d}{dx} [\ln u - \ln v] = \frac{u'}{u} - \frac{v'}{v}.
\]
Now multiply through by \(f(x) = \frac{u}{v}\):
\[
\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{u}{v} \cdot \left(\frac{u'}{u} - \frac{v'}{v}\right).
\]
Simplify:
\[
\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}.
\]
Thus, the quotient rule is derived.

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Let me know if you’d like any further clarifications or details!

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### Related Questions:
1. What is the chain rule, and how does it apply to logarithmic differentiation?
2. Can logarithmic differentiation be used for other functions, such as \(u(x)^v(x)\)?
3. Why does the derivative of a constant term vanish in differentiation?
4. How does the product rule extend to more than two functions (e.g., \(u \cdot v \cdot w\))?
5. Can you derive higher-order derivatives for \(f(x) = u \cdot v\)?

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### Tip:
Logarithmic differentiation simplifies the process for products, quotients, and powers by breaking down complex expressions into sums or differences.

Prove the product and quotient rules for derivatives using logarithmic differentiation and the chain rule.

Understand how to derive the product and quotient rules for differentiation using logarithmic differentiation and the chain rule. Includes detailed examples and solutions for advanced high school math students.

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