1. Find the derivative of $$y = \ln(x^3 - 2x + 1)$$. Step 1: Use the chain rule for the natural logarithm function. Step 2: Derivative of $$\ln u = \frac{1}{u} \cdot u'$$ where $$u = x^3 - 2x + 1$$. Step 3: Compute $$u' = 3x^2 - 2$$. Step 4: So, $$y' = \frac{3x^2 - 2}{x^3 - 2x + 1}$$. 2. Find the derivative of $$y = \ln(\tan x + 1)$$. Step 1: Use the chain rule with $$u = \tan x + 1$$. Step 2: Derivative $$u' = \sec^2 x$$. Step 3: Thus, $$y' = \frac{\sec^2 x}{\tan x + 1}$$. 3. Find the derivative of $$y = x^2 \ln(x^2 + 1)$$. Step 1: Use the product rule: $$y' = (x^2)' \ln(x^2 + 1) + x^2 \cdot \frac{d}{dx} \ln(x^2 + 1)$$. Step 2: Compute first term derivative: $$2x \ln(x^2 + 1)$$. Step 3: Compute second term derivative: $$x^2 \cdot \frac{2x}{x^2 + 1} = \frac{2x^3}{x^2 + 1}$$. Step 4: Final derivative $$y' = 2x \ln(x^2 + 1) + \frac{2x^3}{x^2 + 1}$$. 4. Find the derivative of $$y = \ln(x^2 + 3)$$. Step 1: Use chain rule: derivative of $$\ln u = \frac{1}{u} u'$$ with $$u = x^2 + 3$$. Step 2: Derivative $$u' = 2x$$. Step 3: Final derivative $$y' = \frac{2x}{x^2 + 3}$$. 5. Find the derivative of $$y = \ln(\sin x) \cos x$$. Step 1: Use product rule: $$y' = (\ln(\sin x))' \cos x + \ln(\sin x) (-\sin x)$$. Step 2: Derivative of $$\ln(\sin x)$$ is $$\frac{1}{\sin x} \cos x = \cot x$$. Step 3: Substitute: $$y' = \cot x \cos x - \ln(\sin x) \sin x$$. 6. Find the derivative of $$y = e^{x^2} - 1$$. Step 1: Derivative of constant $$-1$$ is 0. Step 2: Use chain rule for $$e^{x^2}$$: $$2x e^{x^2}$$. Step 3: Final derivative $$y' = 2x e^{x^2}$$. 7. Find the derivative of $$y = e^x \cos x$$. Step 1: Use product rule. Step 2: Derivative: $$e^x \cos x + e^x (-\sin x) = e^x(\cos x - \sin x)$$. 8. Find the derivative of $$y = x e^{\sin x}$$. Step 1: Product rule. Step 2: Derivative of $$x$$ is 1. Step 3: Derivative of $$e^{\sin x}$$ is $$e^{\sin x} \cos x$$. Step 4: Final derivative: $$e^{\sin x} + x e^{\sin x} \cos x = e^{\sin x} (1 + x \cos x)$$. 9. Find the derivative of $$y = e^{x + 4}$$. Step 1: Use chain rule. Step 2: Derivative of $$x+4$$ is 1. Step 3: So $$y' = e^{x+4}$$. 10. Find the derivative of $$y = e^x \tan x$$. Step 1: Use product rule. Step 2: Derivative of $$e^x$$ is $$e^x$$. Step 3: Derivative of $$\tan x$$ is $$\sec^2 x$$. Step 4: Final derivative $$y' = e^x \tan x + e^x \sec^2 x = e^x (\tan x + \sec^2 x)$$. 11. Find the derivative of $$y = x^3 \sin(2x)$$. Step 1: Use product rule. Step 2: Derivative of $$x^3$$ is $$3x^2$$. Step 3: Derivative of $$\sin(2x)$$ is $$2 \cos(2x)$$. Step 4: Final derivative $$y' = 3x^2 \sin(2x) + x^3 \cdot 2 \cos(2x) = 3x^2 \sin(2x) + 2x^3 \cos(2x)$$. 12. Find the derivative of $$y = \cos^2(3x)$$ aka $$(\cos(3x))^2$$. Step 1: Use chain rule and power rule. Step 2: Derivative $$2 \cos(3x)(-\sin(3x)) \cdot 3 = -6 \cos(3x) \sin(3x)$$. 13. Find the derivative of $$y = \frac{\tan x}{\ln x}$$. Step 1: Use quotient rule. Step 2: Derivatives: $$\tan x' = \sec^2 x$$, $$\ln x' = \frac{1}{x}$$. Step 3: Quotient rule formula: $$y' = \frac{(\sec^2 x)(\ln x) - (\tan x)\left(\frac{1}{x}\right)}{(\ln x)^2}$$. 14. Find the derivative of $$y = e^x \cos(\ln x)$$. Step 1: Use product rule. Step 2: Derivative $$e^x$$ is $$e^x$$. Step 3: Derivative of $$\cos(\ln x) = -\sin(\ln x) \cdot \frac{1}{x}$$. Step 4: Final derivative: $$y' = e^x \cos(\ln x) - e^x \frac{\sin(\ln x)}{x} = e^x \left( \cos(\ln x) - \frac{\sin(\ln x)}{x} \right)$$. 15. Find the derivative of $$y = \sin(e^x)$$. Step 1: Use chain rule. Step 2: Derivative of $$\sin u = \cos u \cdot u'$$ where $$u = e^x$$. Step 3: Derivative $$u' = e^x$$. Step 4: Final derivative $$y' = \cos(e^x) e^x$$.