y = $x^{x\cos x}$ and z = $\frac{x^2+1}{x^2-1}$ Consider y = $x^{x\cos x}$ Taking log on both sides, log y = log $(x^{x\cos x})$ log y = x cos x log x Differentiating with respect to x, $\frac{1}{y} \frac{dy}{dx}$ = (x cos x) $\frac{1}{x}$ + log x $\frac{d}{dx}$ + log x $\frac{d}{dx}$ (x cos x) $\frac{1}{y} \frac{dy}{dx}$ = cos x + log x (cos x - x sin x) $\frac{dy}{dx}$ = y (cos x + log x [cos x - x sin x)] $\frac{dy}{dx}$ = $x^{x\cos x}$ [cos x + log x (cos x - x sin x)] … (1) Consider z = $\frac{x^2+1}{x^2-1}$ Differentiating with respect to x, $\frac{dz}{dx}$ = = $\frac{(x^2-1)(2x) - (x^2+1)(2x)}{(x^2-1)^2}$ = $\frac{2x^3-2x-2x^3-2x}{(x^2-1)^2}$ = $\frac{-4x}{(x^2-1)^2}$ ... (2) Adding (1) and (2): $\frac{d}{dx} \left\{ x^{x\cos x} + \frac{x^2+1}{x^2-1} \right\}$ = $\frac{dy}{dx} + \frac{dz}{dx}$ = $x^{x\cos x}$ [cos x + log x (cos x – x sin x)] – $\frac{x}{(x^2-1)^2}$ OR x = a(θ - sinθ) , y = a(1 + cosθ) Differentiating x and y w.r.t. θ, $\frac{dx}{d\theta}$ = a (1 - cos θ) ... (1) $\frac{dy}{d\theta}$ = - a sin θ ... (2) Dividing (2) by (1), $\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$ = $\frac{-a \sin \theta}{a(1-\cos \theta)}$ ⇒ $\frac{dy}{dx}$ = $\frac{-\sin \theta}{1-\cos \theta}$ ⇒ $\frac{dy}{dx}$ = $\frac{-2 \sin\frac{\theta}{2} \cos\frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}$ ⇒ $\frac{dy}{dx}$ = $\frac{-\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}$ ⇒ $\frac{dy}{dx}$ = = - cot $\frac{\theta}{2}$ ... (3) Differentiating w.r.t. x, $\frac{d}{dx} \left( \frac{dy}{dx} \right)$ = $\frac{d}{d\theta} \left( \frac{dy}{dx} \right) \times \frac{d\theta}{dx}$ ⇒ $\frac{d^2y}{dx^2}$ = $\frac{d}{d\theta} \left( \frac{dy}{dx} \right) \times \frac{d\theta}{dx}$ ⇒ $\frac{d^2y}{dx^2}$ = $\frac{d}{d\theta} \left( -\cot \frac{\theta}{2} \right) \times \frac{d\theta}{dx}$ [from equation (3)] $\frac{d^2y}{dx^2}$ = - $\left( -\csc^2 \frac{\theta}{2} \times \frac{1}{2} \right) \times \frac{d\theta}{dx}$ = $\frac{1}{2} \csc^2 \frac{\theta}{2} \times \frac{1}{\frac{dx}{d\theta}}$ = $\frac{1}{2} \csc^2 \frac{\theta}{2} \times \frac{1}{a(1-\cos \theta)}$ ... [from equation (1)] = $\frac{\csc^2 \frac{\theta}{2}}{2a(1-\cos \theta)}$ = $\frac{\csc^2 \frac{\theta}{2}}{2a\left(2 \sin^2 \frac{\theta}{2}\right)}$ = $\frac{1}{4a} \times \csc^2 \frac{\theta}{2}$