Derivatives of Functions in Parametric Forms Questions in English

(A) Given $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$.
Let $u = \sec \theta$ and $v = \cos \theta = \frac{1}{u}$.
Then $x = u - \frac{1}{u}$ and $y = u^n - \frac{1}{u^n}$.
We know that $x^2 = (u - \frac{1}{u})^2 = u^2 + \frac{1}{u^2} - 2$,so $x^2 + 4 = u^2 + \frac{1}{u^2} + 2 = (u + \frac{1}{u})^2$.
Thus,$u + \frac{1}{u} = \sqrt{x^2 + 4}$.
Similarly,$y^2 + 4 = (u^n - \frac{1}{u^n})^2 + 4 = u^{2n} + \frac{1}{u^{2n}} - 2 + 4 = u^{2n} + \frac{1}{u^{2n}} + 2 = (u^n + \frac{1}{u^n})^2$.
Thus,$u^n + \frac{1}{u^n} = \sqrt{y^2 + 4}$.
Using the identity $(u^n - \frac{1}{u^n})^2 = (u^n + \frac{1}{u^n})^2 - 4$,we have $y^2 = (y^2 + 4) - 4$.
Now,differentiate $x$ and $y$ with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$.
$\frac{dy}{d\theta} = n \sec^{n-1} \theta (\sec \theta \tan \theta) + n \cos^{n-1} \theta \sin \theta = n \tan \theta (\sec^n \theta + \cos^n \theta)$.
Then $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{n(\sec^n \theta + \cos^n \theta)}{\sec \theta + \cos \theta} = \frac{n \sqrt{y^2 + 4}}{\sqrt{x^2 + 4}}$.
Squaring both sides,we get ${\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{{n^2(y^2 + 4)}}{{x^2 + 4}}$.

(B) Given $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$.
Squaring both sides,we get $x^2 = 2^{\csc^{-1} t}$ and $y^2 = 2^{\sec^{-1} t}$.
Taking the product,$x^2 y^2 = 2^{\csc^{-1} t + \sec^{-1} t}$.
Since $\csc^{-1} t + \sec^{-1} t = \frac{\pi}{2}$ for $|t| \ge 1$,we have $x^2 y^2 = 2^{\pi/2}$,which is a constant.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(2^{\pi/2})$
$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$
$2xy(y + x \frac{dy}{dx}) = 0$
Since $x, y \neq 0$,we have $y + x \frac{dy}{dx} = 0$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(C) Given,$y = 3 \sin \theta \cos \theta = \frac{3}{2} \sin 2\theta$.
$\frac{dy}{d\theta} = \frac{3}{2} \cdot 2 \cos 2\theta = 3 \cos 2\theta$.
Given $x = e^{\theta} \sin \theta$.
$\frac{dx}{d\theta} = e^{\theta} \sin \theta + e^{\theta} \cos \theta = e^{\theta} (\sin \theta + \cos \theta)$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3 \cos 2\theta}{e^{\theta} (\sin \theta + \cos \theta)}$.
The tangent is parallel to the $x-$axis when $\frac{dy}{dx} = 0$,which implies $3 \cos 2\theta = 0$.
$\cos 2\theta = 0 \Rightarrow 2\theta = \frac{\pi}{2}$ (since $0 \leq \theta \leq \pi$,$0 \leq 2\theta \leq 2\pi$).
$2\theta = \frac{\pi}{2}$ or $2\theta = \frac{3\pi}{2}$.
$\theta = \frac{\pi}{4}$ or $\theta = \frac{3\pi}{4}$.
Checking the denominator $e^{\theta} (\sin \theta + \cos \theta)$ at $\theta = \frac{3\pi}{4}$: $\sin \frac{3\pi}{4} + \cos \frac{3\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$.
Since the denominator becomes zero at $\theta = \frac{3\pi}{4}$,the derivative is undefined.
Thus,the only valid solution is $\theta = \frac{\pi}{4}$.

(D) Given $x = \sqrt{a^{\sin^{-1} t}}$. Taking natural logarithm on both sides,$2 \ln x = (\sin^{-1} t) \ln a$.
Differentiating with respect to $t$,$\frac{2}{x} \frac{dx}{dt} = \frac{\ln a}{\sqrt{1-t^2}}$.
Thus,$\frac{dx}{dt} = \frac{x \ln a}{2\sqrt{1-t^2}}$.
Similarly,$y = \sqrt{a^{\cos^{-1} t}} \Rightarrow 2 \ln y = (\cos^{-1} t) \ln a$.
Differentiating with respect to $t$,$\frac{2}{y} \frac{dy}{dt} = -\frac{\ln a}{\sqrt{1-t^2}}$.
Thus,$\frac{dy}{dt} = -\frac{y \ln a}{2\sqrt{1-t^2}}$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-y \ln a / (2\sqrt{1-t^2})}{x \ln a / (2\sqrt{1-t^2})} = -\frac{y}{x}$.
Therefore,$1 + \left( \frac{dy}{dx} \right)^2 = 1 + \left( -\frac{y}{x} \right)^2 = 1 + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2}$.

(D) Given $x = 3 \tan t$ and $y = 3 \sec t.$
Differentiating with respect to $t$:
$\frac{dx}{dt} = 3 \sec^2 t$
$\frac{dy}{dt} = 3 \sec t \tan t$
Using the chain rule:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \sec t \tan t}{3 \sec^2 t} = \frac{\tan t}{\sec t} = \sin t$
Now,differentiating $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\sin t) = \cos t \cdot \frac{dt}{dx}$
Since $\frac{dx}{dt} = 3 \sec^2 t,$ then $\frac{dt}{dx} = \frac{1}{3 \sec^2 t}.$
Substituting this:
$\frac{d^2y}{dx^2} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos^3 t}{3}$
At $t = \frac{\pi}{4},$ $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}.$
Therefore,$\frac{d^2y}{dx^2} = \frac{(1/\sqrt{2})^3}{3} = \frac{1}{3 \cdot 2\sqrt{2}} = \frac{1}{6\sqrt{2}}.$

(D) Given $x = 2 \sin \theta - \sin 2 \theta$ and $y = 2 \cos \theta - \cos 2 \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 2 \cos \theta - 2 \cos 2 \theta = 2(\cos \theta - (2 \cos^2 \theta - 1)) = 2(1 + \cos \theta - 2 \cos^2 \theta) = 2(1 - \cos \theta)(1 + 2 \cos \theta)$.
Alternatively,using sum-to-product formulas: $\frac{dx}{d\theta} = 2(\cos \theta - \cos 2 \theta) = 4 \sin \frac{3\theta}{2} \sin \frac{\theta}{2}$.
$\frac{dy}{d\theta} = -2 \sin \theta + 2 \sin 2 \theta = 2(\sin 2 \theta - \sin \theta) = 2(2 \sin \theta \cos \theta - \sin \theta) = 2 \sin \theta (2 \cos \theta - 1)$.
Using sum-to-product: $\frac{dy}{d\theta} = 2(\sin 2 \theta - \sin \theta) = 4 \cos \frac{3\theta}{2} \sin \frac{\theta}{2}$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{4 \cos(3\theta/2) \sin(\theta/2)}{4 \sin(3\theta/2) \sin(\theta/2)} = \cot \frac{3\theta}{2}$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{d\theta} \left( \cot \frac{3\theta}{2} \right) \cdot \frac{d\theta}{dx} = -\frac{3}{2} \csc^2 \frac{3\theta}{2} \cdot \frac{1}{dx/d\theta}$.
At $\theta = \pi$,$\frac{dx}{d\theta} = 2 \cos \pi - 2 \cos 2 \pi = 2(-1) - 2(1) = -4$.
$\frac{d^2y}{dx^2} = -\frac{3}{2} \csc^2 \frac{3\pi}{2} \cdot \frac{1}{-4} = -\frac{3}{2} (-1)^2 \cdot \left( -\frac{1}{4} \right) = \frac{3}{8}$.

(A) Given the parametric equations:
$x = a \cos \theta$
$y = a \sin \theta$
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta) = -a \sin \theta$
Next,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = \frac{d}{d\theta}(a \sin \theta) = a \cos \theta$
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
Substitute the values:
$\frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta}$
Simplify the expression:
$\frac{dy}{dx} = -\frac{\cos \theta}{\sin \theta} = -\cot \theta$

(A) Given the parametric equations:
$x = at^2$ and $y = 2at$.
Differentiating $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$.
Differentiating $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}$.

(A) Given parametric equations are $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a(1 + \cos \theta)$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a(0 - (-\sin \theta)) = a \sin \theta$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 + \cos \theta)}$.
Using trigonometric identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$:
$\frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = \tan \frac{\theta}{2}$.

(A) Given the parametric equations:
$x = 2at^{2}$
$y = at^{4}$
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(2at^{2}) = 2a \cdot (2t) = 4at$
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(at^{4}) = a \cdot (4t^{3}) = 4at^{3}$
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4at^{3}}{4at} = t^{2}$

(A) Given equations are $x = a \cos \theta$ and $y = b \cos \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta) = -a \sin \theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = \frac{d}{d\theta}(b \cos \theta) = -b \sin \theta$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-b \sin \theta}{-a \sin \theta} = \frac{b}{a}$.
Thus,the value of $\frac{dy}{dx}$ is $\frac{b}{a}$.

(A) Given equations are $x = \sin t$ and $y = \cos 2t$.
Differentiating $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t$.
Differentiating $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(\cos 2t) = -\sin 2t \cdot \frac{d}{dt}(2t) = -2 \sin 2t$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2 \sin 2t}{\cos t}$.
Using the trigonometric identity $\sin 2t = 2 \sin t \cos t$:
$\frac{dy}{dx} = \frac{-2(2 \sin t \cos t)}{\cos t} = -4 \sin t$.

(A) Given the parametric equations $x = 4t$ and $y = \frac{4}{t}$.
Differentiating $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(4t) = 4$.
Differentiating $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}\left(\frac{4}{t}\right) = 4 \cdot \frac{d}{dt}(t^{-1}) = 4 \cdot (-1 \cdot t^{-2}) = -\frac{4}{t^2}$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{4}{t^2}}{4} = -\frac{1}{t^2}$.

(A) Given equations are $x = \cos \theta - \cos 2\theta$ and $y = \sin \theta - \sin 2\theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = \frac{d}{d\theta}(\cos \theta) - \frac{d}{d\theta}(\cos 2\theta) = -\sin \theta - (-2 \sin 2\theta) = 2 \sin 2\theta - \sin \theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\sin 2\theta) = \cos \theta - 2 \cos 2\theta$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\cos \theta - 2 \cos 2\theta}{2 \sin 2\theta - \sin \theta}$.

(A) Given equations are $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a[\frac{d}{d\theta}(\theta) - \frac{d}{d\theta}(\sin \theta)] = a(1-\cos \theta)$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a[\frac{d}{d\theta}(1) + \frac{d}{d\theta}(\cos \theta)] = a(0 - \sin \theta) = -a \sin \theta$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-a \sin \theta}{a(1-\cos \theta)}$.
Using trigonometric identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1-\cos \theta = 2 \sin^2 \frac{\theta}{2}$:
$\frac{dy}{dx} = \frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}} = -\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = -\cot \frac{\theta}{2}$.

(A) Given equations are $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$.
First,find $\frac{dx}{dt}$ using the quotient rule:
$\frac{dx}{dt} = \frac{\sqrt{\cos 2t} \cdot \frac{d}{dt}(\sin^3 t) - \sin^3 t \cdot \frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t}$
$= \frac{\sqrt{\cos 2t} \cdot 3\sin^2 t \cos t - \sin^3 t \cdot \frac{1}{2\sqrt{\cos 2t}} \cdot (-2\sin 2t)}{\cos 2t}$
$= \frac{3\cos 2t \sin^2 t \cos t + \sin^3 t \sin 2t}{(\cos 2t)^{3/2}}$
Next,find $\frac{dy}{dt}$ using the quotient rule:
$\frac{dy}{dt} = \frac{\sqrt{\cos 2t} \cdot \frac{d}{dt}(\cos^3 t) - \cos^3 t \cdot \frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t}$
$= \frac{\sqrt{\cos 2t} \cdot 3\cos^2 t (-\sin t) - \cos^3 t \cdot \frac{1}{2\sqrt{\cos 2t}} \cdot (-2\sin 2t)}{\cos 2t}$
$= \frac{-3\cos 2t \cos^2 t \sin t + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}}$
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3\cos 2t \cos^2 t \sin t + \cos^3 t \sin 2t}{3\cos 2t \sin^2 t \cos t + \sin^3 t \sin 2t}$
Substitute $\sin 2t = 2\sin t \cos t$:
$= \frac{\sin t \cos t [-3\cos 2t \cos t + 2\cos^3 t]}{\sin t \cos t [3\cos 2t \sin t + 2\sin^3 t]}$
$= \frac{-3(2\cos^2 t - 1)\cos t + 2\cos^3 t}{3(1 - 2\sin^2 t)\sin t + 2\sin^3 t}$
$= \frac{-6\cos^3 t + 3\cos t + 2\cos^3 t}{3\sin t - 6\sin^3 t + 2\sin^3 t} = \frac{-4\cos^3 t + 3\cos t}{3\sin t - 4\sin^3 t}$
Using identities $\cos 3t = 4\cos^3 t - 3\cos t$ and $\sin 3t = 3\sin t - 4\sin^3 t$:
$= \frac{-(\cos 3t)}{\sin 3t} = -\cot 3t$.

(A) Given equations are $x=a\left(\cos t+\log \tan \frac{t}{2}\right)$ and $y=a \sin t$.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}(\log \tan \frac{t}{2}) \right]$
$= a \left[ -\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2} \right]$
$= a \left[ -\sin t + \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \cdot \frac{1}{\cos^2 \frac{t}{2}} \cdot \frac{1}{2} \right]$
$= a \left[ -\sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}} \right]$
$= a \left[ -\sin t + \frac{1}{\sin t} \right] = a \left( \frac{1 - \sin^2 t}{\sin t} \right) = a \frac{\cos^2 t}{\sin t}$.
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = a \frac{d}{dt}(\sin t) = a \cos t$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} = \frac{\cos t \cdot \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = \tan t$.

(A) Given equations are $x = a \sec \theta$ and $y = b \tan \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a \sec \theta \tan \theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = b \sec^2 \theta$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta}$.
Simplifying the expression:
$\frac{dy}{dx} = \frac{b}{a} \cdot \frac{\sec \theta}{\tan \theta} = \frac{b}{a} \cdot \frac{1/\cos \theta}{\sin \theta / \cos \theta} = \frac{b}{a} \cdot \frac{1}{\sin \theta} = \frac{b}{a} \csc \theta$.

(A) Given equations are $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$.
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a \left[ \frac{d}{d\theta}(\cos \theta) + \frac{d}{d\theta}(\theta \sin \theta) \right]$
$= a [-\sin \theta + (\theta \cos \theta + \sin \theta \cdot 1)]$
$= a [-\sin \theta + \theta \cos \theta + \sin \theta] = a \theta \cos \theta$.
Next,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a \left[ \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\theta \cos \theta) \right]$
$= a [\cos \theta - (\theta(-\sin \theta) + \cos \theta \cdot 1)]$
$= a [\cos \theta + \theta \sin \theta - \cos \theta] = a \theta \sin \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta$.

Given equations are $x = \sqrt{a^{\sin^{-1}t}}$ and $y = \sqrt{a^{\cos^{-1}t}}$.
This can be written as $x = a^{\frac{1}{2}\sin^{-1}t}$ and $y = a^{\frac{1}{2}\cos^{-1}t}$.
Taking the natural logarithm on both sides for $x$:
$\ln x = \frac{1}{2}\sin^{-1}t \ln a$.
Differentiating with respect to $t$:
$\frac{1}{x} \frac{dx}{dt} = \frac{\ln a}{2} \cdot \frac{1}{\sqrt{1-t^2}} \implies \frac{dx}{dt} = \frac{x \ln a}{2\sqrt{1-t^2}}$.
Similarly,for $y$:
$\ln y = \frac{1}{2}\cos^{-1}t \ln a$.
Differentiating with respect to $t$:
$\frac{1}{y} \frac{dy}{dt} = \frac{\ln a}{2} \cdot \left(-\frac{1}{\sqrt{1-t^2}}\right) \implies \frac{dy}{dt} = -\frac{y \ln a}{2\sqrt{1-t^2}}$.
Now,using the chain rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{-\frac{y \ln a}{2\sqrt{1-t^2}}}{\frac{x \ln a}{2\sqrt{1-t^2}}} = -\frac{y}{x}$.
Hence,$\frac{dy}{dx} = -\frac{y}{x}$ is proved.

(D) Differentiating $x = a \sin^{3} t$ and $y = b \cos^{3} t$ with respect to $t$,we get:
$\frac{dx}{dt} = 3a \sin^{2} t \cos t$ and $\frac{dy}{dt} = -3b \cos^{2} t \sin t$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3b \cos^{2} t \sin t}{3a \sin^{2} t \cos t} = -\frac{b}{a} \cot t$.
At $t = \frac{\pi}{2}$,the slope is $\left. \frac{dy}{dx} \right|_{t = \frac{\pi}{2}} = -\frac{b}{a} \cot \frac{\pi}{2} = -\frac{b}{a} (0) = 0$.
At $t = \frac{\pi}{2}$,the coordinates of the point are $x = a \sin^{3}(\frac{\pi}{2}) = a(1)^{3} = a$ and $y = b \cos^{3}(\frac{\pi}{2}) = b(0)^{3} = 0$. So the point is $(a, 0)$.
The equation of the tangent line at $(a, 0)$ with slope $m = 0$ is $y - y_{1} = m(x - x_{1})$.
$y - 0 = 0(x - a)$,which simplifies to $y = 0$.

(A) Given the parametric equations of the curve are $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 3a \cos^{2} \theta (-\sin \theta) = -3a \cos^{2} \theta \sin \theta$
$\frac{dy}{d\theta} = 3a \sin^{2} \theta (\cos \theta) = 3a \sin^{2} \theta \cos \theta$
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^{2} \theta \cos \theta}{-3a \cos^{2} \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
At $\theta = \frac{\pi}{4}$,the slope of the tangent is $m_{T} = -\tan(\frac{\pi}{4}) = -1$.
The slope of the normal $m_{N}$ is given by $-\frac{1}{m_{T}}$.
Therefore,$m_{N} = -\frac{1}{-1} = 1$.

(A) Given the parametric equations of the curve are $x = 1 - a \sin \theta$ and $y = b \cos^{2} \theta$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = -a \cos \theta$
$\frac{dy}{d\theta} = b \cdot 2 \cos \theta \cdot (-\sin \theta) = -2b \sin \theta \cos \theta$
Now,the slope of the tangent $\frac{dy}{dx}$ is:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2b \sin \theta \cos \theta}{-a \cos \theta} = \frac{2b}{a} \sin \theta$
At $\theta = \frac{\pi}{2}$,the slope of the tangent is:
$m_{tangent} = \frac{2b}{a} \sin(\frac{\pi}{2}) = \frac{2b}{a} \cdot 1 = \frac{2b}{a}$
The slope of the normal is the negative reciprocal of the slope of the tangent:
$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{2b/a} = -\frac{a}{2b}$

(A) Given the curve $x = \cos t$ and $y = \sin t$.
At $t = \frac{\pi}{4}$,$x = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $y = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
The point is $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
Now,$\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{-\sin t} = -\cot t$.
At $t = \frac{\pi}{4}$,the slope $m = -\cot(\frac{\pi}{4}) = -1$.
The equation of the tangent is $y - \frac{1}{\sqrt{2}} = -1(x - \frac{1}{\sqrt{2}})$,which simplifies to $x + y - \sqrt{2} = 0$.
The slope of the normal is $m' = -\frac{1}{m} = -\frac{1}{-1} = 1$.
The equation of the normal is $y - \frac{1}{\sqrt{2}} = 1(x - \frac{1}{\sqrt{2}})$,which simplifies to $x - y = 0$.

(A) Given $y = a^{t+\frac{1}{t}}$ and $x = \left(t+\frac{1}{t}\right)^{a}$.
First,differentiate $y$ with respect to $t$ using the chain rule:
$\frac{dy}{dt} = \frac{d}{dt}\left(a^{t+\frac{1}{t}}\right) = a^{t+\frac{1}{t}} \cdot \log a \cdot \frac{d}{dt}\left(t+\frac{1}{t}\right) = a^{t+\frac{1}{t}} \cdot \log a \cdot \left(1 - \frac{1}{t^2}\right)$.
Next,differentiate $x$ with respect to $t$ using the power rule:
$\frac{dx}{dt} = \frac{d}{dt}\left(t+\frac{1}{t}\right)^{a} = a\left(t+\frac{1}{t}\right)^{a-1} \cdot \frac{d}{dt}\left(t+\frac{1}{t}\right) = a\left(t+\frac{1}{t}\right)^{a-1} \cdot \left(1 - \frac{1}{t^2}\right)$.
Now,find $\frac{dy}{dx}$ using the parametric form formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{a^{t+\frac{1}{t}} \cdot \log a \cdot \left(1 - \frac{1}{t^2}\right)}{a\left(t+\frac{1}{t}\right)^{a-1} \cdot \left(1 - \frac{1}{t^2}\right)}$.
Canceling the common term $\left(1 - \frac{1}{t^2}\right)$ (for $t \neq \pm 1$):
$\frac{dy}{dx} = \frac{a^{t+\frac{1}{t}} \log a}{a\left(t+\frac{1}{t}\right)^{a-1}}$.

(A) Let $u(x) = \sin ^{2} x$ and $v(x) = e^{\cos x}$.
We need to find the derivative of $u$ with respect to $v$,which is given by $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
First,differentiate $u(x)$ with respect to $x$ using the chain rule:
$\frac{du}{dx} = \frac{d}{dx}(\sin ^{2} x) = 2 \sin x \cos x$.
Next,differentiate $v(x)$ with respect to $x$ using the chain rule:
$\frac{dv}{dx} = \frac{d}{dx}(e^{\cos x}) = e^{\cos x} \cdot \frac{d}{dx}(\cos x) = e^{\cos x} \cdot (-\sin x) = -\sin x e^{\cos x}$.
Now,substitute these into the formula for $\frac{du}{dv}$:
$\frac{du}{dv} = \frac{2 \sin x \cos x}{-\sin x e^{\cos x}}$.
Assuming $\sin x \neq 0$,we can cancel $\sin x$ from the numerator and denominator:
$\frac{du}{dv} = \frac{2 \cos x}{-e^{\cos x}} = -2 \cos x e^{-\cos x}$.

(A) Given $y=12(1-\cos t)$ and $x=10(t-\sin t)$.
Differentiating $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}[10(t-\sin t)] = 10(1-\cos t)$.
Differentiating $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}[12(1-\cos t)] = 12(0 - (-\sin t)) = 12\sin t$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{12\sin t}{10(1-\cos t)}$.
Using trigonometric identities $\sin t = 2\sin \frac{t}{2} \cos \frac{t}{2}$ and $1-\cos t = 2\sin^2 \frac{t}{2}$:
$\frac{dy}{dx} = \frac{12(2\sin \frac{t}{2} \cos \frac{t}{2})}{10(2\sin^2 \frac{t}{2})} = \frac{12}{10} \cot \frac{t}{2} = \frac{6}{5} \cot \frac{t}{2}$.

(D) Given that,$x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t).$
First,we find $\frac{dx}{dt}$:
$\frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}(t \sin t) \right] = a [-\sin t + \sin t + t \cos t] = at \cos t.$
Next,we find $\frac{dy}{dt}$:
$\frac{dy}{dt} = a \left[ \frac{d}{dt}(\sin t) - \frac{d}{dt}(t \cos t) \right] = a [\cos t - (\cos t - t \sin t)] = at \sin t.$
Now,calculate $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at \sin t}{at \cos t} = \tan t.$
Finally,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan t) = \sec^2 t \cdot \frac{dt}{dx}.$
Since $\frac{dx}{dt} = at \cos t,$ we have $\frac{dt}{dx} = \frac{1}{at \cos t}.$
Therefore,$\frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{1}{at \cos t} = \frac{\sec^3 t}{at}.$

(D) Given $x = 2 \sqrt{2} \cos t \sqrt{\sin 2t}$ and $y = 2 \sqrt{2} \sin t \sqrt{\sin 2t}$.
First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = 2 \sqrt{2} [-\sin t \sqrt{\sin 2t} + \cos t \cdot \frac{1}{2\sqrt{\sin 2t}} \cdot 2 \cos 2t] = \frac{2 \sqrt{2} [-\sin t \sin 2t + \cos t \cos 2t]}{\sqrt{\sin 2t}} = \frac{2 \sqrt{2} \cos 3t}{\sqrt{\sin 2t}}$.
Similarly,$\frac{dy}{dt} = 2 \sqrt{2} [\cos t \sqrt{\sin 2t} + \sin t \cdot \frac{1}{2\sqrt{\sin 2t}} \cdot 2 \cos 2t] = \frac{2 \sqrt{2} [\cos t \sin 2t + \sin t \cos 2t]}{\sqrt{\sin 2t}} = \frac{2 \sqrt{2} \sin 3t}{\sqrt{\sin 2t}}$.
Thus,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin 3t}{\cos 3t} = \tan 3t$.
At $t = \frac{\pi}{4}$,$\frac{dy}{dx} = \tan(\frac{3\pi}{4}) = -1$.
Now,find $\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan 3t) = \frac{d}{dt}(\tan 3t) \cdot \frac{dt}{dx} = 3 \sec^2 3t \cdot \frac{\sqrt{\sin 2t}}{2 \sqrt{2} \cos 3t} = \frac{3 \sec^3 3t \sqrt{\sin 2t}}{2 \sqrt{2}}$.
At $t = \frac{\pi}{4}$,$\sec 3t = \sec(\frac{3\pi}{4}) = -\sqrt{2}$,so $\sec^3 3t = -2 \sqrt{2}$.
$\frac{d^2y}{dx^2} = \frac{3(-2 \sqrt{2}) \sqrt{\sin(\pi/2)}}{2 \sqrt{2}} = -3$.
Finally,$\frac{1 + (dy/dx)^2}{d^2y/dx^2} = \frac{1 + (-1)^2}{-3} = \frac{2}{-3} = -\frac{2}{3}$.

(C) Given the parametric equations of the curve: $x = \theta + \sin \theta$ and $y = 1 + \cos \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 1 + \cos \theta$ and $\frac{dy}{d\theta} = -\sin \theta$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-\sin \theta}{1 + \cos \theta}$.
At $\theta = \frac{\pi}{2}$,the point $(x, y)$ is:
$x = \frac{\pi}{2} + \sin \frac{\pi}{2} = \frac{\pi}{2} + 1$ and $y = 1 + \cos \frac{\pi}{2} = 1$.
The slope of the tangent at $\theta = \frac{\pi}{2}$ is $m_t = \frac{-\sin(\pi/2)}{1 + \cos(\pi/2)} = \frac{-1}{1 + 0} = -1$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
The equation of the normal at $(\frac{\pi}{2} + 1, 1)$ is:
$y - 1 = 1(x - (\frac{\pi}{2} + 1))$
$y - 1 = x - \frac{\pi}{2} - 1$
$x - y - \frac{\pi}{2} = 0$,which can be written as $2x - 2y - \pi = 0$.

(B) Given the parametric equations of the curve: $x=\theta+\sin \theta$ and $y=1+\cos \theta$.
First,find the derivative $\frac{dy}{dx}$ with respect to $\theta$:
$\frac{dx}{d\theta} = 1+\cos \theta$ and $\frac{dy}{d\theta} = -\sin \theta$.
Thus,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-\sin \theta}{1+\cos \theta}$.
At $\theta = \frac{\pi}{2}$,the slope of the tangent is $m_t = \frac{-\sin(\pi/2)}{1+\cos(\pi/2)} = \frac{-1}{1+0} = -1$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
Now,find the point $(x, y)$ at $\theta = \frac{\pi}{2}$:
$x = \frac{\pi}{2} + \sin(\frac{\pi}{2}) = \frac{\pi}{2} + 1$ and $y = 1 + \cos(\frac{\pi}{2}) = 1 + 0 = 1$.
The equation of the normal is $(y - y_1) = m_n(x - x_1)$:
$(y - 1) = 1(x - (1 + \frac{\pi}{2}))$.
$y - 1 = x - 1 - \frac{\pi}{2}$.
$x - y - \frac{\pi}{2} = 0$,which simplifies to $2x - 2y - \pi = 0$.

(D) Given the parametric equations $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$.
At $t=4$,$x=\sqrt{4}=2$ and $y=4-\frac{1}{\sqrt{4}}=4-\frac{1}{2}=\frac{7}{2}$.
We find the derivative $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
$\frac{dx}{dt} = \frac{1}{2\sqrt{t}}$ and $\frac{dy}{dt} = 1 + \frac{1}{2t\sqrt{t}}$.
At $t=4$,$\frac{dx}{dt} = \frac{1}{4}$ and $\frac{dy}{dt} = 1 + \frac{1}{16} = \frac{17}{16}$.
Thus,the slope of the tangent $m_T = \frac{17/16}{1/4} = \frac{17}{4}$.
The slope of the normal $m_N = -\frac{1}{m_T} = -\frac{4}{17}$.
The equation of the normal is $y - \frac{7}{2} = -\frac{4}{17}(x - 2)$.
Multiplying by $34$,we get $34y - 119 = -8(x - 2) \implies 34y - 119 = -8x + 16$.
Rearranging gives $8x + 34y = 135$.

(C) Given the parametric equations of the curve are $x = \sqrt{t}$ and $y = t - \frac{1}{\sqrt{t}}$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(t^{1/2}) = \frac{1}{2\sqrt{t}}$
$\frac{dy}{dt} = \frac{d}{dt}(t - t^{-1/2}) = 1 - (-\frac{1}{2})t^{-3/2} = 1 + \frac{1}{2t^{3/2}}$
Now,the slope of the tangent $\frac{dy}{dx}$ is given by $\frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{1 + \frac{1}{2t^{3/2}}}{\frac{1}{2\sqrt{t}}} = (1 + \frac{1}{2t^{3/2}}) \times (2\sqrt{t}) = 2\sqrt{t} + \frac{2\sqrt{t}}{2t^{3/2}} = 2\sqrt{t} + \frac{1}{t} = \frac{2t\sqrt{t} + 1}{t}$
At $t=4$,the slope of the tangent is:
$\left(\frac{dy}{dx}\right)_{t=4} = \frac{2(4)\sqrt{4} + 1}{4} = \frac{2(4)(2) + 1}{4} = \frac{16+1}{4} = \frac{17}{4}$
The slope of the normal is the negative reciprocal of the slope of the tangent:
$\text{Slope of normal} = -\frac{1}{(\frac{dy}{dx})_{t=4}} = -\frac{1}{17/4} = -\frac{4}{17}$

(D) Given the parametric equations of the curve are $x=t^{2}-1$ and $y=t^{2}-t$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2t-1$.
The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t-1}{2t}$.
Since the tangent is parallel to the $X$-axis,its slope must be zero:
$\frac{2t-1}{2t} = 0$.
This implies $2t-1 = 0$,which gives $t = \frac{1}{2}$.

(A) Given the position coordinates: $x = a + bt - ct^2$ and $y = at + bt^2$.
To find the acceleration,we first find the velocity components by differentiating with respect to time $t$:
$v_x = \frac{dx}{dt} = b - 2ct$
$v_y = \frac{dy}{dt} = a + 2bt$
Now,we find the acceleration components by differentiating velocity with respect to time $t$:
$a_x = \frac{dv_x}{dt} = -2c$
$a_y = \frac{dv_y}{dt} = 2b$
The resultant acceleration $a$ is given by the magnitude of the acceleration vector:
$a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-2c)^2 + (2b)^2}$
$a = \sqrt{4c^2 + 4b^2} = 2 \sqrt{c^2 + b^2} \text{ unit/s}^2$.

(A) Given: $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$.
Squaring and adding both equations:
$x^{2} + y^{2} = (a \sin t - b \cos t)^{2} + (a \cos t + b \sin t)^{2}$
$x^{2} + y^{2} = a^{2} \sin^{2} t + b^{2} \cos^{2} t - 2ab \sin t \cos t + a^{2} \cos^{2} t + b^{2} \sin^{2} t + 2ab \sin t \cos t$
$x^{2} + y^{2} = a^{2}(\sin^{2} t + \cos^{2} t) + b^{2}(\cos^{2} t + \sin^{2} t) = a^{2} + b^{2}$.
Differentiating $x^{2} + y^{2} = a^{2} + b^{2}$ with respect to $x$:
$2x + 2y \frac{dy}{dx} = 0 \implies y \frac{dy}{dx} = -x$.
Differentiating again with respect to $x$:
$y \frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2} = -1$.
Substitute $\frac{dy}{dx} = -\frac{x}{y}$:
$y \frac{d^{2}y}{dx^{2}} + \left(-\frac{x}{y}\right)^{2} = -1$
$y \frac{d^{2}y}{dx^{2}} + \frac{x^{2}}{y^{2}} = -1$.
Multiplying by $y^{2}$:
$y^{3} \frac{d^{2}y}{dx^{2}} + x^{2} = -y^{2}$
$y^{3} \frac{d^{2}y}{dx^{2}} + x^{2} + y^{2} = 0$.

(D) Given $x=\sec \theta-\cos \theta$ and $y=\sec ^{10} \theta-\cos ^{10} \theta$.
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$
$\frac{dy}{d\theta} = 10 \sec^9 \theta (\sec \theta \tan \theta) - 10 \cos^9 \theta (-\sin \theta) = 10 \sec^{10} \theta \tan \theta + 10 \cos^9 \theta \sin \theta$
$= 10 \tan \theta (\sec^{10} \theta + \cos^{10} \theta)$
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{10 \tan \theta (\sec^{10} \theta + \cos^{10} \theta)}{\tan \theta (\sec \theta + \cos \theta)} = \frac{10 (\sec^{10} \theta + \cos^{10} \theta)}{\sec \theta + \cos \theta}$
Squaring both sides:
$(\frac{dy}{dx})^2 = \frac{100 (\sec^{10} \theta + \cos^{10} \theta)^2}{(\sec \theta + \cos \theta)^2}$
Using the identity $(a+b)^2 = (a-b)^2 + 4ab$:
$(\frac{dy}{dx})^2 = \frac{100 [(\sec^{10} \theta - \cos^{10} \theta)^2 + 4 \sec^{10} \theta \cos^{10} \theta]}{(\sec \theta - \cos \theta)^2 + 4 \sec \theta \cos \theta}$
Since $\sec \theta \cos \theta = 1$,we have:
$(\frac{dy}{dx})^2 = \frac{100 (y^2 + 4)}{x^2 + 4}$
Thus,$(x^2+4)(\frac{dy}{dx})^2 = 100(y^2+4)$.
Comparing with the given equation,$k = 100$.

(B) Let $t = \tan^{-1} x$. Then $u = \frac{t}{t+1}$ and $v = \tan^{-1}(t)$.
We need to find $\frac{du}{dv} = \frac{du/dt}{dv/dt}$.
First,differentiate $u$ with respect to $t$ using the quotient rule:
$\frac{du}{dt} = \frac{(t+1)(1) - t(1)}{(t+1)^2} = \frac{1}{(t+1)^2}$.
Next,differentiate $v$ with respect to $t$:
$\frac{dv}{dt} = \frac{1}{1+t^2}$.
Now,calculate $\frac{du}{dv}$:
$\frac{du}{dv} = \frac{du/dt}{dv/dt} = \frac{1/(t+1)^2}{1/(1+t^2)} = \frac{1+t^2}{(t+1)^2}$.
Substituting $t = \tan^{-1} x$ back into the expression:
$\frac{du}{dv} = \frac{1 + (\tan^{-1} x)^2}{(1 + \tan^{-1} x)^2}$.

(C) Given $y=f\left(\frac{2 x+3}{3-2 x}\right)$.
Applying the chain rule,we have $\frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$.
Using the quotient rule for $\frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$:
$\frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right) = \frac{(3-2 x)(2) - (2 x+3)(-2)}{(3-2 x)^2} = \frac{6-4 x+4 x+6}{(3-2 x)^2} = \frac{12}{(3-2 x)^2}$.
Since $f^{\prime}(x)=\sin (\log x)$,then $f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) = \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)$.
Thus,$\frac{d y}{d x} = \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right) \cdot \frac{12}{(3-2 x)^2}$.
At $x=1$,the argument of the function is $\frac{2(1)+3}{3-2(1)} = \frac{5}{1} = 5$.
The derivative at $x=1$ is $\sin (\log 5) \cdot \frac{12}{(3-2)^2} = 12 \sin (\log 5)$.

(B) Given $x = \sin \theta$ and $y = \sin^3 \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = \cos \theta$
$\frac{dy}{d\theta} = 3 \sin^2 \theta \cos \theta$
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3 \sin^2 \theta \cos \theta}{\cos \theta} = 3 \sin^2 \theta$
Next,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(3 \sin^2 \theta) = \frac{d}{d\theta}(3 \sin^2 \theta) \cdot \frac{d\theta}{dx}$
$\frac{d^2 y}{dx^2} = (6 \sin \theta \cos \theta) \cdot \frac{1}{\cos \theta} = 6 \sin \theta$
Finally,evaluate at $\theta = \frac{\pi}{2}$:
$\frac{d^2 y}{dx^2} = 6 \sin(\frac{\pi}{2}) = 6(1) = 6$.

(C) Let $u = \sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}}$ and $v = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}$.
Given $u = v$.
For $u$,we have $u = \sqrt{y-u} \implies u^2 = y-u \implies y = u^2+u$.
For $v$,we have $v = \sqrt{x+v} \implies v^2 = x+v \implies x = v^2-v$.
Since $u = v$,we can write $y = u^2+u$ and $x = u^2-u$.
Now,differentiate $y$ and $x$ with respect to $u$:
$\frac{dy}{du} = 2u+1$ and $\frac{dx}{du} = 2u-1$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/du}{dx/du} = \frac{2u+1}{2u-1}$.
From $y = u^2+u$,we have $u^2+u-y = 0$. Solving for $u$ using the quadratic formula: $u = \frac{-1 \pm \sqrt{1+4y}}{2}$.
Alternatively,express the result in terms of $x$ and $y$:
Since $y-x = (u^2+u) - (u^2-u) = 2u$,we have $u = \frac{y-x}{2}$.
Substituting $u$ into $\frac{dy}{dx} = \frac{2u+1}{2u-1}$:
$\frac{dy}{dx} = \frac{2(\frac{y-x}{2})+1}{2(\frac{y-x}{2})-1} = \frac{y-x+1}{y-x-1}$.

(D) Let $\sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}} = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} = z$.
Since the expressions are infinite,we can write:
$y - z = z^2 \Rightarrow y = z^2 + z$
$x + z = z^2 \Rightarrow x = z^2 - z$
Now,differentiate both with respect to $z$:
$\frac{dy}{dz} = 2z + 1$
$\frac{dx}{dz} = 2z - 1$
Using the chain rule,$\frac{dy}{dx} = \frac{dy/dz}{dx/dz} = \frac{2z+1}{2z-1}$.
From the equations $y = z^2 + z$ and $x = z^2 - z$,we subtract them:
$y - x = (z^2 + z) - (z^2 - z) = 2z$.
Substituting $2z = y - x$ into the derivative expression:
$\frac{dy}{dx} = \frac{(y-x) + 1}{(y-x) - 1} = \frac{y-x+1}{y-x-1}$.

(C) Given $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 3a \cos^2 \theta (-\sin \theta) = -3a \cos^2 \theta \sin \theta$.
$\frac{dy}{d\theta} = 3a \sin^2 \theta (\cos \theta) = 3a \sin^2 \theta \cos \theta$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
Then,$\left(\frac{dy}{dx}\right)^2 = (-\tan \theta)^2 = \tan^2 \theta$.
Finally,$\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = |\sec \theta|$.

(A) Given $x = a \sin 2t (1 + \cos 2t) = a \sin 2t + a \sin 2t \cos 2t = a \sin 2t + \frac{a}{2} \sin 4t$.
Differentiating with respect to $t$: $\frac{dx}{dt} = 2a \cos 2t + 2a \cos 4t = 2a (\cos 2t + \cos 4t)$.
Given $y = b \cos 2t (1 - \cos 2t) = b \cos 2t - b \cos^2 2t$.
Using $\cos^2 2t = \frac{1 + \cos 4t}{2}$,we have $y = b \cos 2t - \frac{b}{2} (1 + \cos 4t)$.
Differentiating with respect to $t$: $\frac{dy}{dt} = -2b \sin 2t + b \sin 4t = -2b \sin 2t + 2b \sin 2t \cos 2t = -2b \sin 2t (1 - \cos 2t)$.
Using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2b \sin 2t (1 - \cos 2t)}{2a (\cos 2t + \cos 4t)} = \frac{-b \sin 2t (1 - \cos 2t)}{a \cos 2t (1 + 2 \cos 2t - 1)} = \frac{-b \sin 2t (1 - \cos 2t)}{a \cos 2t (2 \cos 2t)} = \frac{-b (2 \sin t \cos t) (2 \sin^2 t)}{a (2 \cos^2 t - 1) (2 \cos 2t)} = \frac{b}{a} \tan t$.

(C) Given $x = \sin \theta$ and $y = \sin^3 \theta$.
First,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{d\theta} = 3 \sin^2 \theta \cos \theta$
$\frac{dx}{d\theta} = \cos \theta$
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3 \sin^2 \theta \cos \theta}{\cos \theta} = 3 \sin^2 \theta$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx} (3 \sin^2 \theta) = \frac{d}{d\theta} (3 \sin^2 \theta) \cdot \frac{d\theta}{dx}$.
Since $\frac{dx}{d\theta} = \cos \theta$,then $\frac{d\theta}{dx} = \frac{1}{\cos \theta}$.
$\frac{d^2 y}{dx^2} = (6 \sin \theta \cos \theta) \cdot \frac{1}{\cos \theta} = 6 \sin \theta$.
At $\theta = \frac{\pi}{6}$:
$\frac{d^2 y}{dx^2} = 6 \sin \left( \frac{\pi}{6} \right) = 6 \cdot \frac{1}{2} = 3$.

(A) Given $x = \sin t$ and $y = \sin pt$.
First,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = p \cos pt$.
So,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p \cos pt}{\cos t}$.
This implies $\cos t \frac{dy}{dx} = p \cos pt$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx} (\cos t \frac{dy}{dx}) = \frac{d}{dx} (p \cos pt)$.
Using the product rule on the left:
$-\sin t \frac{dt}{dx} \frac{dy}{dx} + \cos t \frac{d^2 y}{dx^2} = -p^2 \sin pt \frac{dt}{dx}$.
Since $\frac{dt}{dx} = \frac{1}{\cos t}$,we substitute:
$-\sin t (\frac{1}{\cos t}) \frac{dy}{dx} + \cos t \frac{d^2 y}{dx^2} = -p^2 \sin pt (\frac{1}{\cos t})$.
Multiply the entire equation by $\cos t$:
$-\sin t \frac{dy}{dx} + \cos^2 t \frac{d^2 y}{dx^2} = -p^2 \sin pt$.
Substitute $x = \sin t$,$\cos^2 t = 1 - x^2$,and $y = \sin pt$:
$-x \frac{dy}{dx} + (1 - x^2) \frac{d^2 y}{dx^2} = -p^2 y$.
Rearranging the terms:
$(1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + p^2 y = 0$.

(D) Given $x = t^2 + t + 1$ and $y = \sin \left(\frac{t \pi}{2}\right) + \cos \left(\frac{t \pi}{2}\right)$.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(t^2 + t + 1) = 2t + 1$.
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}\left(\sin \left(\frac{t \pi}{2}\right) + \cos \left(\frac{t \pi}{2}\right)\right) = \frac{\pi}{2} \cos \left(\frac{t \pi}{2}\right) - \frac{\pi}{2} \sin \left(\frac{t \pi}{2}\right)$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{\pi}{2} \left(\cos \left(\frac{t \pi}{2}\right) - \sin \left(\frac{t \pi}{2}\right)\right)}{2t + 1}$.
At $t=1$:
$\frac{dy}{dx} = \frac{\frac{\pi}{2} \left(\cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{2}\right)\right)}{2(1) + 1} = \frac{\frac{\pi}{2} (0 - 1)}{3} = \frac{-\pi/2}{3} = -\frac{\pi}{6}$.
Thus,the correct option is $D$.

(C) Let $t = \tan \theta$. Then $\theta = \tan^{-1} t$.
For $x$: $x = \tan^{-1} \left\{ \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta} \right\} = \tan^{-1} \left\{ \frac{\sec \theta - 1}{\tan \theta} \right\} = \tan^{-1} \left\{ \frac{1-\cos \theta}{\sin \theta} \right\} = \tan^{-1} \left\{ \tan \frac{\theta}{2} \right\} = \frac{\theta}{2} = \frac{1}{2} \tan^{-1} t$.
Thus,$\frac{dx}{dt} = \frac{1}{2(1+t^2)}$.
For $y$: $y = \cos^{-1} \left\{ \frac{1-\tan^2 \theta}{1+\tan^2 \theta} \right\} = \cos^{-1} (\cos 2\theta) = 2\theta = 2 \tan^{-1} t$.
Thus,$\frac{dy}{dt} = \frac{2}{1+t^2}$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2/(1+t^2)}{1/(2(1+t^2))} = 2 \times 2 = 4$.

(D) Let $u = \sin ^2 x$ and $v = e^{\cos x}$.
First,differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(\sin ^2 x) = 2 \sin x \cdot \cos x$.
Next,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx}(e^{\cos x}) = e^{\cos x} \cdot (-\sin x) = -\sin x \cdot e^{\cos x}$.
Now,find the derivative of $u$ with respect to $v$ using the chain rule:
$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2 \sin x \cdot \cos x}{-\sin x \cdot e^{\cos x}}$.
Canceling $\sin x$ from the numerator and denominator,we get:
$\frac{du}{dv} = \frac{2 \cos x}{-e^{\cos x}} = \frac{-2 \cos x}{e^{\cos x}}$.