1. Problem (a): Find the derivative of $$\tan^{-1}(\sqrt{3}x) + (\tan^{-1}x^2)^2$$.\nStep 1: Use chain rule and derivative of arctan: $$\frac{d}{dx}[\tan^{-1}u] = \frac{u'}{1+u^2}$$.\nFor $$\tan^{-1}(\sqrt{3}x)$$, $$u = \sqrt{3}x$$, so $$u' = \sqrt{3}$$.\\ Derivative is $$\frac{\sqrt{3}}{1 + 3x^2}$$.\nStep 2: For $$(\tan^{-1}x^2)^2$$, use chain rule: $$2 \tan^{-1}(x^2) \times \frac{d}{dx}[\tan^{-1}(x^2)]$$.\nDerivative inside is $$\frac{2x}{1 + x^4}$$.\nSo total derivative is $$2 \tan^{-1}(x^2) \cdot \frac{2x}{1+x^4} = \frac{4x \tan^{-1}(x^2)}{1+x^4}$$.\nStep 3: Add these for final answer: $$\frac{\sqrt{3}}{1 + 3x^2} + \frac{4x \tan^{-1}(x^2)}{1+x^4}$$.\n\n2. Problem (b): Derivative of $$\csc^{-1}(x^{-1}) + \cot^{-1}x$$.\nStep 1: Use derivative $$\frac{d}{dx}[\csc^{-1}u] = - \frac{u'}{|u|\sqrt{u^2-1}}$$ and $$\frac{d}{dx}[\cot^{-1}x] = -\frac{1}{1+x^2}$$.\n\nStep 2: For $$\csc^{-1}(x^{-1})$$, $$u = x^{-1}$$, so $$u' = -x^{-2}$$.\nDerivative is $$- \frac{-x^{-2}}{|x^{-1}| \sqrt{(x^{-1})^2 -1}} = \frac{x^{-2}}{(1/|x|) \sqrt{x^{-2} -1}} = \frac{x^{-2} |x|}{\sqrt{x^{-2} -1}} = \frac{|x|}{x^{2} \sqrt{x^{-2} -1}}$$.\\ Since $$\sqrt{x^{-2}-1} = \frac{\sqrt{1 - x^2}}{|x|}$$, this simplifies to $$\frac{|x|}{x^{2}} \cdot \frac{|x|}{\sqrt{1-x^2}} = \frac{1}{x \sqrt{1-x^2}}$$ for $$|x|>0$$ and $$|x| \neq 1$$.\nStep 3: Add derivative of $$\cot^{-1}x$$ which is $$-\frac{1}{1+x^2}$$.\nFinal derivative: $$\frac{1}{x \sqrt{1-x^{2}}} - \frac{1}{1 + x^{2}}$$.\n\n3. Problem (c): Derivative of $$\frac{\sin^{-1} x}{\sin x} + \sqrt{x^2+1} \sec^{-1} x$$.\nStep 1: Use quotient rule for $$\frac{\sin^{-1} x}{\sin x}$$: $$\frac{(\sin^{-1} x)' \sin x - \sin^{-1} x (\sin x)'}{\sin^2 x} = \frac{\frac{1}{\sqrt{1-x^2}} \sin x - \sin^{-1} x \cos x}{\sin^2 x}$$.\nStep 2: Derivative of $$\sqrt{x^2+1} \sec^{-1} x$$ is product rule: $$(\sqrt{x^2+1})' \sec^{-1} x + \sqrt{x^2+1} (\sec^{-1} x)'$$\nwhere $$(\sqrt{x^2+1})' = \frac{x}{\sqrt{x^2+1}}$$ and $$(\sec^{-1} x)' = \frac{1}{|x| \sqrt{x^2 - 1}}$$ for $$|x| > 1$$.\nSo derivative is $$\frac{x}{\sqrt{x^2 +1}} \sec^{-1} x + \frac{\sqrt{x^2 +1}}{|x| \sqrt{x^2 -1}}$$.\nStep 3: Final derivative combines both parts.\n\n4. Problem (d): Derivative of $$\tan^{-1}(\cos x) + \cos^{-1}(\sin^{-1} x)$$.\nStep 1: Use chain rule: $$\frac{d}{dx}[\tan^{-1}(\cos x)] = \frac{-\sin x}{1 + \cos^2 x}$$.\nStep 2: For $$\cos^{-1}(\sin^{-1} x)$$, derivative is $$- \frac{1}{\sqrt{1 - (\sin^{-1} x)^2}} \cdot \frac{1}{\sqrt{1 - x^2}}$$ by chain rule since derivative of $$\sin^{-1} x$$ is $$1/\sqrt{1-x^2}$$.\n\n5. Problem (3): Given $$y = \frac{\sin^{-1} x}{\sqrt{1-x^2}}$$, show $$ (1 - x^2) y'' = 3 x y' + y$$.\nStep 1: Compute $$y'$$ using quotient rule and derivative of $$\sin^{-1} x$$.\nStep 2: Compute $$y''$$ by differentiating $$y'$$.\nStep 3: Substitute $$y, y', y''$$ into equation and verify equality holds by simplifying both sides.\n\n6. Problem (4): Given $$f(x) = \sin^{-1} \frac{x-1}{x+1}$$ and $$g(x) = 2 \tan^{-1}(\sqrt{x})$$, show $$f'(x) = g'(x)$$.\nStep 1: Find $$f'(x)$$ by chain rule: derivative of $$\sin^{-1} u$$ is $$\frac{u'}{\sqrt{1-u^2}}$$ with $$u = \frac{x-1}{x+1}$$. \nStep 2: Compute derivative of $$g(x)$$: $$g'(x) = 2 \cdot \frac{1}{1+ (\sqrt{x})^2} \cdot \frac{1}{2 \sqrt{x}} = \frac{1}{\sqrt{x} (1 + x)}$$.\nStep 3: Simplify both derivatives to verify they are equal.\n\n7. Problem (5): $$y = \frac{x^2}{2} \cos^{-1} x + \frac{1}{4} \sin^{-1} x - \frac{x}{4} \sqrt{1-x^2}$$, show $$y' = x \cos^{-1} x$$.\nStep 1: Differentiate each term with product and chain rules.\nStep 2: Simplify derivatives carefully to reduce to $$x \cos^{-1} x$$.\n\n8. Problem (6): Tangent line to $$\tan^{-1}(xy) = \sin^{-1}(x + y)$$ at (0,0).\nStep 1: Differentiate both sides implicitly.\nStep 2: At (0,0), substitute and solve for $$\frac{dy}{dx}$$.\nStep 3: Use point-slope form to write tangent line equation.\n\n9. Problem (7): Points on $$x^2 + y^2 = 25$$ where tangent slope = 1/2.\nStep 1: Differentiate implicitly: $$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$$.\nStep 2: Set slope $$= \frac{1}{2} = -\frac{x}{y} \Rightarrow y = -2x$$.\nStep 3: Substitute in circle: $$x^2 + (-2x)^2 = 25 \Rightarrow x^2 + 4x^2 = 25 \Rightarrow 5x^2 = 25 \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5}$$.\nStep 4: Compute corresponding $$y = -2x$$, points: $$(\sqrt{5}, -2\sqrt{5}), (-\sqrt{5}, 2\sqrt{5})$$.