1. **Problem:** Find the derivative of $f(x) = (\ln x)^2$. 2. **Formula:** Use the chain rule: If $f(x) = [g(x)]^2$, then $f'(x) = 2g(x)g'(x)$. 3. **Work:** Here, $g(x) = \ln x$, so $g'(x) = \frac{1}{x}$. 4. **Derivative:** $$f'(x) = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}$$ --- 1. **Problem:** Find the derivative of $h(x) = \ln \sqrt{x^2 + 1}$. 2. **Rewrite:** $h(x) = \ln (x^2 + 1)^{1/2} = \frac{1}{2} \ln (x^2 + 1)$. 3. **Formula:** Derivative of $\ln u = \frac{u'}{u}$. 4. **Work:** $$h'(x) = \frac{1}{2} \cdot \frac{2x}{x^2 + 1} = \frac{x}{x^2 + 1}$$ --- 1. **Problem:** Find the derivative of $f(x) = \frac{\ln(\sin x)}{x}$. 2. **Formula:** Use the quotient rule: If $f = \frac{u}{v}$, then $$f' = \frac{u'v - uv'}{v^2}$$ 3. **Work:** - $u = \ln(\sin x)$, so $u' = \frac{\cos x}{\sin x} = \cot x$. - $v = x$, so $v' = 1$. 4. **Derivative:** $$f'(x) = \frac{\cot x \cdot x - \ln(\sin x) \cdot 1}{x^2} = \frac{x \cot x - \ln(\sin x)}{x^2}$$ --- 1. **Problem:** Find the derivative of $h(x) = \tan x - \frac{\ln x}{3x}$. 2. **Formula:** Derivative of $\tan x$ is $\sec^2 x$. 3. **For** $\frac{\ln x}{3x}$, use quotient rule with $u = \ln x$, $v = 3x$: - $u' = \frac{1}{x}$ - $v' = 3$ 4. **Derivative of second term:** $$\left(\frac{\ln x}{3x}\right)' = \frac{\frac{1}{x} \cdot 3x - \ln x \cdot 3}{(3x)^2} = \frac{3 - 3 \ln x}{9x^2} = \frac{1 - \ln x}{3x^2}$$ 5. **Final derivative:** $$h'(x) = \sec^2 x - \frac{1 - \ln x}{3x^2}$$