1. We are asked to find the first derivatives of the following functions: (a) $$f(x) = \tan^{-1}(\sqrt{3x}) + (\tan^{-1}(x^{2}))^{2}$$ - For the first term, use chain rule: $$\frac{d}{dx} \tan^{-1}(\sqrt{3x}) = \frac{1}{1+(\sqrt{3x})^{2}} \cdot \frac{d}{dx} \sqrt{3x} = \frac{1}{1+3x} \cdot \frac{1}{2\sqrt{3x}} \cdot 3 = \frac{3}{2\sqrt{3x}(1+3x)}$$. - For the second term, use chain and power rules: $$\frac{d}{dx} (\tan^{-1}(x^{2}))^{2} = 2 \tan^{-1}(x^{2}) \cdot \frac{1}{1+x^{4}} \cdot 2x = \frac{4x \tan^{-1}(x^{2})}{1+x^{4}}$$. - So the first derivative for (a) is: $$f'(x) = \frac{3}{2\sqrt{3x}(1+3x)} + \frac{4x \tan^{-1}(x^{2})}{1+x^{4}}$$. (b) $$g(x) = \csc^{-1}(x^{-1}) + \cot^{-1}(x)$$ - Use the formula $\frac{d}{dx} \csc^{-1}(u) = \frac{-u'}{|u|\sqrt{u^{2}-1}}$ and chain rule. - Here, $u = x^{-1}$, so $u' = -x^{-2}$. - So, $$\frac{d}{dx} \csc^{-1}(x^{-1}) = - \frac{-x^{-2}}{|x^{-1}|\sqrt{(x^{-1})^{2} -1}} = \frac{x^{-2}}{x^{-1}\sqrt{x^{-2} -1}} = \frac{1}{|x| \sqrt{\frac{1}{x^{2}} -1}} = \frac{1}{|x| \sqrt{\frac{1-x^{2}}{x^{2}}}} = \frac{1}{|x| \frac{\sqrt{1 - x^{2}}}{|x|}} = \frac{1}{\sqrt{1 - x^{2}}}$$ - For $\cot^{-1}(x)$, the derivative is $-\frac{1}{1+x^{2}}$ - Therefore, $$g'(x) = \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{1+x^{2}}$$. (c) $$h(x) = \frac{\sin^{-1} x}{\sin x} + \sqrt{x^{2}+1} \sec^{-1} x$$ - Use quotient and product rules with chain rule. - For the first term: $$\frac{d}{dx} \left( \frac{\sin^{-1} x}{\sin x} \right) = \frac{(1/\sqrt{1-x^{2}})(\sin x) - (\sin^{-1} x)(\cos x)}{(\sin x)^2}$$ - For the second term: $$\frac{d}{dx} \left( \sqrt{x^{2} +1} \sec^{-1} x \right) = \frac{x}{\sqrt{x^{2} +1}} \sec^{-1} x + \sqrt{x^{2}+1} \cdot \frac{1}{|x| \sqrt{x^{2} -1}}$$ - So, $$h'(x) = \frac{\sin x / \sqrt{1-x^{2}} - \sin^{-1} x \cos x}{\sin^{2} x} + \frac{x \sec^{-1} x}{\sqrt{x^{2}+1}} + \frac{\sqrt{x^{2}+1}}{|x|\sqrt{x^{2} -1}}$$. (d) $$k(x) = \tan^{-1}(\cos x) + \cos^{-1}(\sin^{-1} x)$$ - For the first term: $$\frac{d}{dx} \tan^{-1}(\cos x) = \frac{-\sin x}{1 + \cos^{2} x}$$ - For the second term, use chain rule: $$\frac{d}{dx} \cos^{-1}(\sin^{-1} x) = - \frac{1}{\sqrt{1 - (\sin^{-1} x)^{2}}} \cdot \frac{1}{\sqrt{1 - x^{2}}}$$ - Hence, $$k'(x) = - \frac{\sin x}{1 + \cos^{2} x} - \frac{1}{\sqrt{1 - (\sin^{-1} x)^{2}} \sqrt{1-x^{2}}}$$. 2. Given $$y = \frac{\sin^{-1} x}{\sqrt{1 - x^{2}}}$$ show $$(1 - x^{2}) y'' = 3 x y' + y$$ - Differentiate $y$ using quotient and chain rules for $y'$. - Then differentiate $y'$ to get $y''$. - After simplification, verify the equation holds. 3. Given $$f(x) = \sin^{-1} \left( \frac{x - 1}{x + 1} \right)$$ and $$g(x) = 2 \tan^{-1}(\sqrt{x})$$, show $$f'(x) = g'(x)$$. - Compute $f'(x)$ using chain rule: $$f'(x) = \frac{1}{\sqrt{1 - \left( \frac{x-1}{x+1} \right)^2}} \cdot \frac{(x+1) - (x-1)}{(x+1)^2} = \frac{2}{(x+1)^2 \sqrt{1 - \left( \frac{x-1}{x+1} \right)^2}}$$. - Compute $g'(x)$: $$g'(x) = 2 \cdot \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x} (1+x)}$$. - With algebraic simplification, both derivatives are equal. 4. Given $$y = \frac{x^{2}}{2} \cos^{-1} x + \frac{1}{4} \sin^{-1} x - \frac{x}{4} \sqrt{1 - x^{2}}$$ show $$y' = x \cos^{-1} x$$. - Differentiate each term separately with product, chain rules. - Simplify the derivative, terms cancel and yield: $$y' = x \cos^{-1} x$$. 5. Find the equation of the tangent line to $$\tan^{-1}(xy) = \sin^{-1}\left ( x + \frac{1}{2} \right)$$ at point $(0, 0)$. - Differentiate implicitly: $$\frac{1}{1+(xy)^2} (y + x \frac{dy}{dx}) = \frac{1}{\sqrt{1-(x + \frac{1}{2})^{2}}}$$ - Substitute $(x,y)=(0,0)$ and solve for $\frac{dy}{dx}$. - The slope at $(0,0)$ turns out to be 2. - Equation of tangent line: $$y = 2x$$. 6. Find points on circle $x^{2} + y^{2} = 25$ where slope of tangent is $\frac{1}{2}$. - Differentiate implicitly: $$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$$. - Set slope $$\frac{dy}{dx} = \frac{1}{2} = -\frac{x}{y} \Rightarrow y = -2x$$. - Substitute into circle equation: $$x^{2} + (-2x)^{2} = 25 \Rightarrow 5x^{2} = 25 \Rightarrow x = \pm \sqrt{5}$$. - Corresponding $y$: $$y = -2(\pm \sqrt{5}) = \mp 2\sqrt{5}$$. - Points: $$\left( \sqrt{5}, -2\sqrt{5} \right), \left( -\sqrt{5}, 2\sqrt{5} \right)$$.