# Chain Rule: Derivatives of the Form $\ln(g(x))$

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### **Example 1:**
Find the derivative of
$$ f(x) = \ln(2x) $$

**Solution:**
- Here, $g(x) = 2x$, so $g'(x) = 2$.
- Using the chain rule:
$$ f'(x) = \frac{1}{2x} \cdot 2 = \frac{2}{2x} = \frac{1}{x} $$

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### **Example 2:**
Find the derivative of
$$ f(x) = \ln(x^2) $$

**Solution:**
- Here, $g(x) = x^2$, so $g'(x) = 2x$.
- Using the chain rule:
$$ f'(x) = \frac{1}{x^2} \cdot 2x = \frac{2x}{x^2} = \frac{2}{x} $$

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### **Example 3:**
Find the derivative of
$$ f(x) = \ln(x^3 + 1) $$

**Solution:**
- Here, $g(x) = x^3 + 1$, so $g'(x) = 3x^2$.
- Using the chain rule:
$$ f'(x) = \frac{1}{x^3 + 1} \cdot 3x^2 = \frac{3x^2}{x^3 + 1} $$

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### **Example 4:**
Find the derivative of
$$ f(x) = \ln(5x^4 - x) $$

**Solution:**
- Here, $g(x) = 5x^4 - x$, so $g'(x) = 20x^3 - 1$.
- Using the chain rule:
$$ f'(x) = \frac{1}{5x^4 - x} \cdot (20x^3 - 1) = \frac{20x^3 - 1}{5x^4 - x} $$

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### **Example 5:**
Find the derivative of
$$ f(x) = \ln\left(\frac{1}{x}\right) $$

**Solution:**
- Here, $g(x) = \frac{1}{x} = x^{-1}$, so $g'(x) = -x^{-2}$.
- Using the chain rule:
$$ f'(x) = \frac{1}{x^{-1}} \cdot (-x^{-2}) = (-x^{-2}) \cdot x = -\frac{1}{x} $$

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### **Example 6:**
Find the derivative of
$$ f(x) = \ln(\sqrt{x}) $$

**Solution:**
- Here, $g(x) = \sqrt{x} = x^{1/2}$, so $g'(x) = \frac{1}{2}x^{-1/2}$.
- Using the chain rule:
$$ f'(x) = \frac{1}{x^{1/2}} \cdot \frac{1}{2}x^{-1/2} = \frac{1}{x^{1/2}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2x} $$

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### **Example 7:**
Find the derivative of
$$ f(x) = \ln(x + 4) $$

**Solution:**
- Here, $g(x) = x + 4$, so $g'(x) = 1$.
- Using the chain rule:
$$ f'(x) = \frac{1}{x + 4} \cdot 1 = \frac{1}{x + 4} $$

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### **Example 8:**
Find the derivative of
$$ f(x) = \ln(3x^5) $$

**Solution:**
- Here, $g(x) = 3x^5$, so $g'(x) = 15x^4$.
- Using the chain rule:
$$ f'(x) = \frac{1}{3x^5} \cdot 15x^4 = \frac{15x^4}{3x^5} = \frac{5}{x} $$

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### **Example 9:**
Find the derivative of
$$ f(x) = \ln(x^2 - 3x + 1) $$

**Solution:**
- Here, $g(x) = x^2 - 3x + 1$, so $g'(x) = 2x - 3$.
- Using the chain rule:
$$ f'(x) = \frac{1}{x^2 - 3x + 1} \cdot (2x - 3) = \frac{2x - 3}{x^2 - 3x + 1} $$

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### **Example 10:**
Find the derivative of
$$ f(x) = \ln(x^{-3}) $$

**Solution:**
- Here, $g(x) = x^{-3}$, so $g'(x) = -3x^{-4}$.
- Using the chain rule:
$$ f'(x) = \frac{1}{x^{-3}} \cdot (-3x^{-4}) = -3x^{-4} \cdot x^3 = -\frac{3}{x} $$

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# Summary
For derivatives of the form $\ln(g(x))$, use the chain rule:
$$ \frac{d}{dx}[\ln(g(x))] = \frac{1}{g(x)} \cdot g'(x) $$

You first differentiate $g(x)$, then divide $g'(x)$ by $g(x)$. Be careful when applying the quotient rule or chain rule within $g(x)$ if it is more complex.