1. **Problem:** Find the derivative of $$y = 5^{2x} \sin^2 x$$. 2. **Formula and rules:** Use the product rule: $$\frac{d}{dx}[u v] = u' v + u v'$$. Also, use the chain rule for $$5^{2x}$$ and $$\sin^2 x$$. Recall: $$\frac{d}{dx} a^{g(x)} = a^{g(x)} \ln a \cdot g'(x)$$ and $$\frac{d}{dx} \sin^2 x = 2 \sin x \cos x$$. 3. **Intermediate work:** - Let $$u = 5^{2x}$$, then $$u' = 5^{2x} \ln 5 \cdot 2 = 2 \ln 5 \cdot 5^{2x}$$. - Let $$v = \sin^2 x$$, then $$v' = 2 \sin x \cos x$$. 4. **Apply product rule:** $$y' = u' v + u v' = 2 \ln 5 \cdot 5^{2x} \sin^2 x + 5^{2x} (2 \sin x \cos x)$$. 5. **Simplify:** $$y' = 5^{2x} \left( 2 \ln 5 \sin^2 x + 2 \sin x \cos x \right)$$. --- 1. **Problem:** Find the derivative of $$y = e^{3x} \ln(3x^3 + 2x^2)$$. 2. **Formula and rules:** Use product rule and chain rule. Recall: $$\frac{d}{dx} e^{g(x)} = e^{g(x)} g'(x)$$ and $$\frac{d}{dx} \ln f(x) = \frac{f'(x)}{f(x)}$$. 3. **Intermediate work:** - Let $$u = e^{3x}$$, then $$u' = e^{3x} \cdot 3 = 3 e^{3x}$$. - Let $$v = \ln(3x^3 + 2x^2)$$, then $$v' = \frac{9x^2 + 4x}{3x^3 + 2x^2}$$. 4. **Apply product rule:** $$y' = u' v + u v' = 3 e^{3x} \ln(3x^3 + 2x^2) + e^{3x} \frac{9x^2 + 4x}{3x^3 + 2x^2}$$. 5. **Simplify:** $$y' = e^{3x} \left( 3 \ln(3x^3 + 2x^2) + \frac{9x^2 + 4x}{3x^3 + 2x^2} \right)$$. --- 1. **Problem:** Find the derivative of $$y = \frac{\sin x - \cos x}{\sin x + \cos x}$$. 2. **Formula and rules:** Use quotient rule: $$\frac{d}{dx} \left( \frac{f}{g} \right) = \frac{f' g - f g'}{g^2}$$. 3. **Intermediate work:** - Let $$f = \sin x - \cos x$$, then $$f' = \cos x + \sin x$$. - Let $$g = \sin x + \cos x$$, then $$g' = \cos x - \sin x$$. 4. **Apply quotient rule:** $$y' = \frac{(\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}$$. 5. **Simplify numerator:** - Expand first term: $$\cos x \sin x + \cos^2 x + \sin^2 x + \sin x \cos x = 2 \sin x \cos x + 1$$ (since $$\sin^2 x + \cos^2 x = 1$$). - Expand second term: $$\sin x \cos x - \sin^2 x - \cos^2 x + \sin x \cos x = 2 \sin x \cos x - 1$$. 6. **Subtract:** $$[2 \sin x \cos x + 1] - [2 \sin x \cos x - 1] = 2$$. 7. **Final derivative:** $$y' = \frac{2}{(\sin x + \cos x)^2}$$. --- 1. **Problem:** Find the derivative of $$y = \arcsin \left( \frac{1 - 2x}{1 + 2x} \right)$$. 2. **Formula and rules:** Use chain rule and derivative of arcsin: $$\frac{d}{dx} \arcsin u = \frac{u'}{\sqrt{1 - u^2}}$$. 3. **Intermediate work:** - Let $$u = \frac{1 - 2x}{1 + 2x}$$. - Compute $$u'$$ using quotient rule: $$u' = \frac{(-2)(1 + 2x) - (1 - 2x)(2)}{(1 + 2x)^2} = \frac{-2 - 4x - 2 + 4x}{(1 + 2x)^2} = \frac{-4}{(1 + 2x)^2}$$. 4. **Compute denominator:** $$1 - u^2 = 1 - \left( \frac{1 - 2x}{1 + 2x} \right)^2 = \frac{(1 + 2x)^2 - (1 - 2x)^2}{(1 + 2x)^2}$$. 5. **Simplify numerator:** $$(1 + 2x)^2 - (1 - 2x)^2 = [(1 + 2x) - (1 - 2x)] \cdot [(1 + 2x) + (1 - 2x)] = (4x)(2) = 8x$$. 6. **So:** $$\sqrt{1 - u^2} = \sqrt{\frac{8x}{(1 + 2x)^2}} = \frac{\sqrt{8x}}{1 + 2x} = \frac{2 \sqrt{2x}}{1 + 2x}$$. 7. **Final derivative:** $$y' = \frac{u'}{\sqrt{1 - u^2}} = \frac{-4/(1 + 2x)^2}{2 \sqrt{2x}/(1 + 2x)} = \frac{-4}{(1 + 2x)^2} \cdot \frac{1 + 2x}{2 \sqrt{2x}} = \frac{-4}{(1 + 2x) 2 \sqrt{2x}} = \frac{-2}{(1 + 2x) \sqrt{2x}}$$. --- 1. **Problem:** Find the derivative of $$y = \operatorname{arccot} \left( \frac{\cos x}{1 - \sin x} \right)$$. 2. **Formula and rules:** Use chain rule and derivative of arccot: $$\frac{d}{dx} \operatorname{arccot} u = - \frac{u'}{1 + u^2}$$. 3. **Intermediate work:** - Let $$u = \frac{\cos x}{1 - \sin x}$$. - Compute $$u'$$ using quotient rule: $$u' = \frac{-\sin x (1 - \sin x) - \cos x (-\cos x)}{(1 - \sin x)^2} = \frac{-\sin x + \sin^2 x + \cos^2 x}{(1 - \sin x)^2} = \frac{-\sin x + 1}{(1 - \sin x)^2}$$ (since $$\sin^2 x + \cos^2 x = 1$$). 4. **Simplify numerator:** $$-\sin x + 1 = 1 - \sin x$$. 5. **So:** $$u' = \frac{1 - \sin x}{(1 - \sin x)^2} = \frac{1}{1 - \sin x}$$. 6. **Compute denominator:** $$1 + u^2 = 1 + \left( \frac{\cos x}{1 - \sin x} \right)^2 = \frac{(1 - \sin x)^2 + \cos^2 x}{(1 - \sin x)^2}$$. 7. **Simplify numerator:** $$(1 - \sin x)^2 + \cos^2 x = 1 - 2 \sin x + \sin^2 x + \cos^2 x = 1 - 2 \sin x + 1 = 2 - 2 \sin x = 2(1 - \sin x)$$. 8. **So:** $$1 + u^2 = \frac{2(1 - \sin x)}{(1 - \sin x)^2} = \frac{2}{1 - \sin x}$$. 9. **Final derivative:** $$y' = - \frac{u'}{1 + u^2} = - \frac{\frac{1}{1 - \sin x}}{\frac{2}{1 - \sin x}} = - \frac{1}{2} = -\frac{1}{2}$$. --- **Final answers:** $$\boxed{y' = 5^{2x} \left( 2 \ln 5 \sin^2 x + 2 \sin x \cos x \right)}$$ $$\boxed{y' = e^{3x} \left( 3 \ln(3x^3 + 2x^2) + \frac{9x^2 + 4x}{3x^3 + 2x^2} \right)}$$ $$\boxed{y' = \frac{2}{(\sin x + \cos x)^2}}$$ $$\boxed{y' = \frac{-2}{(1 + 2x) \sqrt{2x}}}$$ $$\boxed{y' = -\frac{1}{2}}$$