Derivatives of Functions in Parametric Forms Questions in English

(D) Given $y=a \sin ^3 t$ and $x=a \cos ^3 t$.
First,differentiate $y$ with respect to $t$: $\frac{d y}{d t} = 3a \sin ^2 t \cos t$.
Next,differentiate $x$ with respect to $t$: $\frac{d x}{d t} = 3a \cos ^2 t (-\sin t) = -3a \cos ^2 t \sin t$.
Now,find $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{3a \sin ^2 t \cos t}{-3a \cos ^2 t \sin t} = -\frac{\sin t}{\cos t} = -\tan t$.
At $t = \frac{3 \pi}{4}$,$\frac{d y}{d x} = -\tan(\frac{3 \pi}{4}) = -(-1) = 1$.

(B) Given that,$y=e^{\sin ^{-1}(t^{2}-1)}$ and $x=e^{\sec ^{-1}(\frac{1}{t^{2}-1})}$.
Taking natural logarithm on both sides:
$\log y = \sin ^{-1}(t^{2}-1)$
$\log x = \sec ^{-1}(\frac{1}{t^{2}-1}) = \cos ^{-1}(t^{2}-1)$.
Adding the two equations:
$\log y + \log x = \sin ^{-1}(t^{2}-1) + \cos ^{-1}(t^{2}-1)$.
Using the identity $\sin ^{-1}(\theta) + \cos ^{-1}(\theta) = \frac{\pi}{2}$,we get:
$\log(xy) = \frac{\pi}{2}$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\log y + \log x) = \frac{d}{dx}(\frac{\pi}{2})$
$\frac{1}{y} \frac{dy}{dx} + \frac{1}{x} = 0$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(B) Given the parametric equations of the curve:
$x = t^{2} + 3t - 8 \quad (1)$
$y = 2t^{2} - 2t - 5 \quad (2)$
At the point $(2, -1)$,we first find the value of $t$.
From equation $(2)$:
$-1 = 2t^{2} - 2t - 5 \Rightarrow 2t^{2} - 2t - 4 = 0 \Rightarrow t^{2} - t - 2 = 0$
$(t - 2)(t + 1) = 0 \Rightarrow t = 2$ or $t = -1$.
From equation $(1)$:
$2 = t^{2} + 3t - 8 \Rightarrow t^{2} + 3t - 10 = 0$
$(t + 5)(t - 2) = 0 \Rightarrow t = 2$ or $t = -5$.
The common value for $t$ is $t = 2$.
Now,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2t + 3$
$\frac{dy}{dt} = 4t - 2$
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$.
At $t = 2$:
$\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

(D) Given $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$.
Differentiating 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,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^{2} \theta \cos \theta}{-3a \cos^{2} \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
From the given equations:
$\frac{y}{a} = \sin^{3} \theta \Rightarrow \sin \theta = (\frac{y}{a})^{1/3}$.
$\frac{x}{a} = \cos^{3} \theta \Rightarrow \cos \theta = (\frac{x}{a})^{1/3}$.
Substituting these into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{(y/a)^{1/3}}{(x/a)^{1/3}} = -(\frac{y}{x})^{1/3} = -\sqrt[3]{\frac{y}{x}}$.

(B) Let $u = \sin x$ and $v = \log x$.
We need to find the derivative of $u$ with respect to $v$,which is $\frac{du}{dv}$.
Using the chain rule,we have $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
First,find the derivative of $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$.
Next,find the derivative of $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$.
Now,substitute these into the formula: $\frac{du}{dv} = \frac{\cos x}{1/x} = x \cos x$.

(A) Given,$x=e^{\theta} \sin \theta$ and $y=e^{\theta} \cos \theta$.
We differentiate both with respect to $\theta$:
$\frac{dx}{d\theta} = e^{\theta} \sin \theta + e^{\theta} \cos \theta = e^{\theta}(\sin \theta + \cos \theta)$
$\frac{dy}{d\theta} = e^{\theta} \cos \theta - e^{\theta} \sin \theta = e^{\theta}(\cos \theta - \sin \theta)$
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{e^{\theta}(\cos \theta - \sin \theta)}{e^{\theta}(\cos \theta + \sin \theta)} = \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$.
Dividing numerator and denominator by $\cos \theta$,we get $\frac{dy}{dx} = \frac{1 - \tan \theta}{1 + \tan \theta}$.
At $(x, y) = (1, 1)$,we have $x = e^{\theta} \sin \theta = 1$ and $y = e^{\theta} \cos \theta = 1$.
Dividing these,$\frac{y}{x} = \frac{e^{\theta} \cos \theta}{e^{\theta} \sin \theta} = \cot \theta = 1$,so $\tan \theta = 1$.
Substituting $\tan \theta = 1$ into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$.

(C) Let $u = \sin(x^{3})$ and $v = \cos(x^{3})$.
By differentiating both with respect to $x$,we get:
$\frac{du}{dx} = \cos(x^{3}) \cdot 3x^{2}$
$\frac{dv}{dx} = -\sin(x^{3}) \cdot 3x^{2}$
Now,the derivative of $u$ with respect to $v$ is given by the chain rule:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{3x^{2} \cos(x^{3})}{-3x^{2} \sin(x^{3})}$
$\frac{du}{dv} = -\frac{\cos(x^{3})}{\sin(x^{3})} = -\cot(x^{3})$

(D) Given $x = a \sec^{2} \theta$ and $y = a \tan^{2} \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a \cdot 2 \sec \theta \cdot (\sec \theta \tan \theta) = 2a \sec^{2} \theta \tan \theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a \cdot 2 \tan \theta \cdot (\sec^{2} \theta) = 2a \tan \theta \sec^{2} \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a \tan \theta \sec^{2} \theta}{2a \sec^{2} \theta \tan \theta} = 1$.
Finally,differentiate $\frac{dy}{dx}$ with respect to $x$ to find the second derivative:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(1) = 0$.

(D) Given $x = \cos \theta + \log \tan \frac{\theta}{2}$.
On differentiating with respect to $\theta$,we get:
$\frac{dx}{d\theta} = -\sin \theta + \frac{1}{\tan \frac{\theta}{2}} \cdot \sec^2 \frac{\theta}{2} \cdot \frac{1}{2}$
$= -\sin \theta + \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} \cdot \frac{1}{\cos^2 \frac{\theta}{2}} \cdot \frac{1}{2}$
$= -\sin \theta + \frac{1}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = -\sin \theta + \frac{1}{\sin \theta}$
$= \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \quad ...(i)$
Now,$y = \sin \theta$. On differentiating with respect to $\theta$,we get:
$\frac{dy}{d\theta} = \cos \theta \quad ...(ii)$
Therefore,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\cos \theta}{\frac{\cos^2 \theta}{\sin \theta}} = \tan \theta$.
For $\frac{dy}{dx} = 0$,we must have $\tan \theta = 0$.
This implies $\theta = n \pi$ for $n \in Z$.

(C) Given the equations $x = ct$ and $y = \frac{c}{t}$.
We need to find $\frac{dy}{dt}$ at $t = 2$.
Since $y = \frac{c}{t} = c \cdot t^{-1}$,we differentiate with respect to $t$:
$\frac{dy}{dt} = c \cdot (-1) \cdot t^{-2} = -\frac{c}{t^2}$.
At $t = 2$,we substitute the value into the derivative:
$\left. \frac{dy}{dt} \right|_{t=2} = -\frac{c}{(2)^2} = -\frac{c}{4}$.

(B) Given $y=t^2+t^3$ and $x=t-t^4$.
First,find the derivatives with respect to $t$:
$\frac{dy}{dt} = 2t + 3t^2$
$\frac{dx}{dt} = 1 - 4t^3$
Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t + 3t^2}{1 - 4t^3}$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$ using the chain rule:
$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{2t + 3t^2}{1 - 4t^3}\right) \cdot \frac{1}{1 - 4t^3}$.
Using the quotient rule:
$\frac{d}{dt}\left(\frac{2t + 3t^2}{1 - 4t^3}\right) = \frac{(1 - 4t^3)(2 + 6t) - (2t + 3t^2)(-12t^2)}{(1 - 4t^3)^2}$.
Thus,$\frac{d^2y}{dx^2} = \frac{(1 - 4t^3)(2 + 6t) + 12t^2(2t + 3t^2)}{(1 - 4t^3)^3}$.
At $t=1$:
$\left.\frac{d^2y}{dx^2}\right|_{t=1} = \frac{(1 - 4)(2 + 6) + 12(1)^2(2(1) + 3(1)^2)}{(1 - 4)^3} = \frac{(-3)(8) + 12(5)}{(-3)^3} = \frac{-24 + 60}{-27} = \frac{36}{-27} = -\frac{4}{3}$.

(B) Given that $f^{\prime}(x) = \sqrt{2x^2-1}$ and $y = f(x^3)$.
Applying the chain rule to differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = f^{\prime}(x^3) \cdot \frac{d}{dx}(x^3)$
$\frac{dy}{dx} = f^{\prime}(x^3) \cdot 3x^2$
Now,evaluate the derivative at $x=1$:
$\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(1^3) \cdot 3(1)^2$
$= f^{\prime}(1) \cdot 3$
Substitute $x=1$ into the given expression for $f^{\prime}(x)$:
$f^{\prime}(1) = \sqrt{2(1)^2 - 1} = \sqrt{2-1} = \sqrt{1} = 1$
Therefore,$\frac{dy}{dx} = 1 \cdot 3 = 3$.

(A) Given $x = 2 \cos^3 \theta$ and $y = 3 \sin^2 \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = 2 \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -6 \cos^2 \theta \sin \theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = 3 \cdot 2 \sin \theta \cdot \cos \theta = 6 \sin \theta \cos \theta$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{6 \sin \theta \cos \theta}{-6 \cos^2 \theta \sin \theta}$.
Simplifying the expression:
$\frac{dy}{dx} = \frac{1}{-\cos \theta} = -\sec \theta$.

(A) Given $x = \sqrt{2} e^t(\sin t - \cos t)$ and $y = \sqrt{2} e^t(\sin t + \cos t)$.
First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = \sqrt{2} [e^t(\sin t - \cos t) + e^t(\cos t + \sin t)] = \sqrt{2} e^t(2 \sin t) = 2\sqrt{2} e^t \sin t$.
$\frac{dy}{dt} = \sqrt{2} [e^t(\sin t + \cos t) + e^t(\cos t - \sin t)] = \sqrt{2} e^t(2 \cos t) = 2\sqrt{2} e^t \cos t$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\sqrt{2} e^t \cos t}{2\sqrt{2} e^t \sin t} = \cot t$.
Next,find $\frac{d^2 y}{dx^2} = \frac{d}{dx}(\cot t) = \frac{d}{dt}(\cot t) \cdot \frac{dt}{dx} = -\csc^2 t \cdot \frac{1}{dx/dt}$.
Substitute $\frac{dx}{dt} = 2\sqrt{2} e^t \sin t$:
$\frac{d^2 y}{dx^2} = -\csc^2 t \cdot \frac{1}{2\sqrt{2} e^t \sin t} = -\frac{1}{2\sqrt{2} e^t \sin^3 t}$.
At $t = \frac{\pi}{4}$,$\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin^3(\frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}}$.
$\left(\frac{d^2 y}{dx^2}\right)_{t = \pi/4} = -\frac{1}{2\sqrt{2} e^{\pi/4} \cdot (1 / 2\sqrt{2})} = -\frac{1}{e^{\pi/4}} = -e^{-\pi/4}$.

(C) Given,$x=3\left[\sin t-\log \left(\cot \frac{t}{2}\right)\right]$ and $y=6\left[\cos t+\log \left(\tan \frac{t}{2}\right)\right]$.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = 3\left[\cos t - \frac{1}{\cot(t/2)} \cdot (-\csc^2(t/2)) \cdot \frac{1}{2}\right] = 3\left[\cos t + \frac{\csc^2(t/2)}{2 \cot(t/2)}\right] = 3\left[\cos t + \frac{1}{2 \sin(t/2) \cos(t/2)}\right] = 3\left[\cos t + \frac{1}{\sin t}\right] = \frac{3(1+\sin t \cos t)}{\sin t}$.
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = 6\left[-\sin t + \frac{1}{\tan(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}\right] = 6\left[-\sin t + \frac{\sec^2(t/2)}{2 \tan(t/2)}\right] = 6\left[-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}\right] = 6\left[-\sin t + \frac{1}{\sin t}\right] = 6\left(\frac{1-\sin^2 t}{\sin t}\right) = \frac{6 \cos^2 t}{\sin t}$.
Finally,calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{6 \cos^2 t / \sin t}{3(1+\sin t \cos t) / \sin t} = \frac{2 \cos^2 t}{1+\sin t \cos t}$.

(C) Given $u=\sin \left(\frac{x}{y}\right)$,$x=e^t$,and $y=t^2$. Substituting $x$ and $y$ in $u$,we get $u=\sin \left(\frac{e^t}{t^2}\right)$.
Now,differentiating $u$ with respect to $t$ using the chain rule:
$\frac{d u}{d t}=\cos \left(\frac{e^t}{t^2}\right) \cdot \frac{d}{d t}\left(\frac{e^t}{t^2}\right)$
Using the quotient rule for $\frac{d}{d t}\left(\frac{e^t}{t^2}\right) = \frac{t^2 e^t - e^t(2t)}{(t^2)^2} = \frac{t e^t(t-2)}{t^4} = \frac{e^t(t-2)}{t^3}$.
Thus,$\frac{d u}{d t} = \cos \left(\frac{e^t}{t^2}\right) \cdot \frac{e^t(t-2)}{t^3}$.
Squaring both sides:
$\left(\frac{d u}{d t}\right)^2 = \cos^2 \left(\frac{e^t}{t^2}\right) \cdot \frac{e^{2t}(t-2)^2}{t^6}$.
Rearranging the expression $t^6\left(\frac{d u}{d t}\right)^2 \div \left(e^{2 t}(t-2)^2\right)$:
$= \frac{t^6 \cdot \cos^2 \left(\frac{e^t}{t^2}\right) \cdot e^{2t}(t-2)^2}{t^6 \cdot e^{2t}(t-2)^2} = \cos^2 \left(\frac{e^t}{t^2}\right)$.
Since $u = \sin \left(\frac{e^t}{t^2}\right)$,we have $\sin^2 \left(\frac{e^t}{t^2}\right) = u^2$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get $\cos^2 \left(\frac{e^t}{t^2}\right) = 1 - u^2$.

(C) Given,$x=f(\theta)$ and $y=g(\theta)$.
First,find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dx}{d\theta} = f^{\prime}(\theta)$ and $\frac{dy}{d\theta} = g^{\prime}(\theta)$.
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{g^{\prime}(\theta)}{f^{\prime}(\theta)}$.
Now,differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g^{\prime}(\theta)}{f^{\prime}(\theta)} \right) = \frac{d}{d\theta} \left( \frac{g^{\prime}(\theta)}{f^{\prime}(\theta)} \right) \cdot \frac{d\theta}{dx}$.
Using the quotient rule:
$\frac{d}{d\theta} \left( \frac{g^{\prime}(\theta)}{f^{\prime}(\theta)} \right) = \frac{g^{\prime \prime}(\theta)f^{\prime}(\theta) - g^{\prime}(\theta)f^{\prime \prime}(\theta)}{(f^{\prime}(\theta))^2}$.
Since $\frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{f^{\prime}(\theta)}$,we have:
$\frac{d^2y}{dx^2} = \frac{g^{\prime \prime}(\theta)f^{\prime}(\theta) - g^{\prime}(\theta)f^{\prime \prime}(\theta)}{(f^{\prime}(\theta))^2} \cdot \frac{1}{f^{\prime}(\theta)} = \frac{f^{\prime}(\theta)g^{\prime \prime}(\theta) - g^{\prime}(\theta)f^{\prime \prime}(\theta)}{(f^{\prime}(\theta))^3}$.

(A) Given $x=a \cos ^3 \theta$ and $y=a \sin ^3 \theta$.
First,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,find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta$
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-\tan \theta) = \frac{d}{d\theta}(-\tan \theta) \cdot \frac{d\theta}{dx} = -\sec^2 \theta \cdot \frac{1}{dx/d\theta}$
$\frac{d^2y}{dx^2} = -\sec^2 \theta \cdot \frac{1}{-3a \cos^2 \theta \sin \theta} = \frac{1}{3a \cos^4 \theta \sin \theta}$
At $\theta = \frac{\pi}{4}$:
$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$
$\frac{d^2y}{dx^2} = \frac{1}{3a (\frac{1}{\sqrt{2}})^4 (\frac{1}{\sqrt{2}})} = \frac{1}{3a (\frac{1}{4}) (\frac{1}{\sqrt{2}})} = \frac{4\sqrt{2}}{3a}$

(D) Given $x=\sec \theta-\cos \theta$ and $y=\sec ^{10} \theta-\cos ^{10} \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = 10 \sec^9 \theta (\sec \theta \tan \theta) - 10 \cos^9 \theta (-\sin \theta) = 10 \tan \theta (\sec^{10} \theta + \cos^{10} \theta)$.
Thus,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = 10 \frac{\sec^{10} \theta + \cos^{10} \theta}{\sec \theta + \cos \theta}$.
Squaring both sides:
$(\frac{dy}{dx})^2 = 100 \frac{(\sec^{10} \theta + \cos^{10} \theta)^2}{(\sec \theta + \cos \theta)^2}$.
Using the identity $(a+b)^2 = (a-b)^2 + 4ab$:
$(\sec^{10} \theta + \cos^{10} \theta)^2 = (\sec^{10} \theta - \cos^{10} \theta)^2 + 4(\sec^{10} \theta \cos^{10} \theta) = y^2 + 4$.
Similarly,$(\sec \theta + \cos \theta)^2 = (\sec \theta - \cos \theta)^2 + 4 = x^2 + 4$.
Substituting these into the expression for $(\frac{dy}{dx})^2$:
$(\frac{dy}{dx})^2 = 100 \frac{y^2+4}{x^2+4}$.
Therefore,$(x^2+4)(\frac{dy}{dx})^2 = 100(y^2+4)$.
Comparing this with the given equation $(x^2+4)(\frac{dy}{dx})^2 = k(y^2+4)$,we get $k=100$.

(B) Given $x=\sec \theta-\cos \theta$ and $y=\sec ^n \theta-\cos ^n \theta$.
First,differentiate $x$ with respect to $\theta$:
$\frac{d x}{d \theta} = \sec \theta \tan \theta + \sin \theta = \sec \theta \cdot \frac{\sin \theta}{\cos \theta} + \sin \theta = \tan \theta(\sec \theta + \cos \theta)$.
Next,differentiate $y$ with respect to $\theta$:
$\frac{d y}{d \theta} = n \sec ^{n-1} \theta \cdot \sec \theta \tan \theta - n \cos ^{n-1} \theta(-\sin \theta) = n \sec ^n \theta \tan \theta + n \cos ^{n-1} \theta \sin \theta = n \tan \theta(\sec ^n \theta + \cos ^n \theta)$.
Now,find $\frac{d y}{d x} = \frac{d y / d \theta}{d x / d \theta}$:
$\frac{d y}{d x} = \frac{n \tan \theta(\sec ^n \theta + \cos ^n \theta)}{\tan \theta(\sec \theta + \cos \theta)} = \frac{n(\sec ^n \theta + \cos ^n \theta)}{\sec \theta + \cos \theta}$.
Squaring both sides:
$\left(\frac{d y}{d x}\right)^2 = \frac{n^2(\sec ^n \theta + \cos ^n \theta)^2}{(\sec \theta + \cos \theta)^2} = \frac{n^2(\sec ^{2n} \theta + \cos ^{2n} \theta + 2)}{(\sec ^2 \theta + \cos ^2 \theta + 2)}$.
Since $(\sec ^n \theta - \cos ^n \theta)^2 = \sec ^{2n} \theta + \cos ^{2n} \theta - 2$,we have $\sec ^{2n} \theta + \cos ^{2n} \theta = y^2 + 2$.
Substituting this:
$\left(\frac{d y}{d x}\right)^2 = \frac{n^2(y^2 + 2 + 2)}{x^2 + 2 + 2} = \frac{n^2(y^2 + 4)}{x^2 + 4}$.
Therefore,$\frac{d y}{d x} = n \sqrt{\frac{y^2+4}{x^2+4}}$.

(B) Given: $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$.
We know that $\sec \theta = \frac{1}{\cos \theta}$,so $x = \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta}$.
Also,$x^2 + 4 = (\sec \theta - \cos \theta)^2 + 4 = \sec^2 \theta - 2 + \cos^2 \theta + 4 = \sec^2 \theta + 2 + \cos^2 \theta = (\sec \theta + \cos \theta)^2$.
Thus,$\sqrt{x^2 + 4} = \sec \theta + \cos \theta$.
Similarly,$y^2 + 4 = (\sec^n \theta - \cos^n \theta)^2 + 4 = \sec^{2n} \theta - 2 + \cos^{2n} \theta + 4 = \sec^{2n} \theta + 2 + \cos^{2n} \theta = (\sec^n \theta + \cos^n \theta)^2$.
Thus,$\sqrt{y^2 + 4} = \sec^n \theta + \cos^n \theta$.
Now,$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$.
And $\frac{dy}{d\theta} = n \sec^{n-1} \theta (\sec \theta \tan \theta) - n \cos^{n-1} \theta (-\sin \theta) = n \sec^n \theta \tan \theta + n \cos^{n-1} \theta \sin \theta = n \tan \theta (\sec^n \theta + \cos^n \theta)$.
Therefore,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{n \tan \theta (\sec^n \theta + \cos^n \theta)}{\tan \theta (\sec \theta + \cos \theta)} = n \frac{\sec^n \theta + \cos^n \theta}{\sec \theta + \cos \theta} = n \frac{\sqrt{y^2 + 4}}{\sqrt{x^2 + 4}} = n \sqrt{\frac{y^2 + 4}{x^2 + 4}}$.

(B) Given $x = a(\cos t + t \sin t)$.
Differentiating with respect to $t$:
$\frac{dx}{dt} = a(-\sin t + \sin t + t \cos t) = at \cos t$.
Given $y = a(\sin t - t \cos t)$.
Differentiating with respect to $t$:
$\frac{dy}{dt} = a(\cos t - (\cos t - t \sin t)) = a(\cos t - \cos t + t \sin t) = at \sin t$.
Now,calculate $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$:
$= \sqrt{(at \cos t)^2 + (at \sin t)^2}$
$= \sqrt{a^2 t^2 \cos^2 t + a^2 t^2 \sin^2 t}$
$= \sqrt{a^2 t^2 (\cos^2 t + \sin^2 t)}$
$= \sqrt{a^2 t^2 (1)}$
$= at$.

(D) Given $y = t^2 + t^3$ and $x = t - t^4$.
First,find the derivatives with respect to $t$:
$\frac{dy}{dt} = 2t + 3t^2$
$\frac{dx}{dt} = 1 - 4t^3$
Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t + 3t^2}{1 - 4t^3}$.
To find $\frac{d^2y}{dx^2}$,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{2t + 3t^2}{1 - 4t^3} \right) \cdot \frac{dt}{dx} = \frac{d}{dt} \left( \frac{2t + 3t^2}{1 - 4t^3} \right) \cdot \frac{1}{1 - 4t^3}$.
Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$:
$\frac{d}{dt} \left( \frac{2t + 3t^2}{1 - 4t^3} \right) = \frac{(2 + 6t)(1 - 4t^3) - (2t + 3t^2)(-12t^2)}{(1 - 4t^3)^2} = \frac{2 - 8t^3 + 6t - 24t^4 + 24t^3 + 36t^4}{(1 - 4t^3)^2} = \frac{12t^4 + 16t^3 + 6t + 2}{(1 - 4t^3)^2}$.
Thus,$\frac{d^2y}{dx^2} = \frac{12t^4 + 16t^3 + 6t + 2}{(1 - 4t^3)^3}$.
At $t = 1$:
$\frac{d^2y}{dx^2} = \frac{12(1)^4 + 16(1)^3 + 6(1) + 2}{(1 - 4(1)^3)^3} = \frac{12 + 16 + 6 + 2}{(1 - 4)^3} = \frac{36}{(-3)^3} = \frac{36}{-27} = -\frac{4}{3}$.

(C) Given $x = a(1 - \sin t)$ and $y = a(t + \cos t)$.
First,find $\frac{dx}{dt} = -a \cos t$ and $\frac{dy}{dt} = a(1 - \sin t)$.
Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a(1 - \sin t)}{-a \cos t} = \frac{\sin t - 1}{\cos t} = \tan t - \sec t$.
Now,differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan t - \sec t) = \frac{d}{dt}(\tan t - \sec t) \cdot \frac{dt}{dx}$.
$\frac{d^2y}{dx^2} = (\sec^2 t - \sec t \tan t) \cdot \frac{1}{-a \cos t} = \frac{\sec t(\sec t - \tan t)}{-a \cos t} = \frac{\frac{1}{\cos t}(\frac{1 - \sin t}{\cos t})}{-a \cos t} = \frac{1 - \sin t}{-a \cos^3 t} = \frac{\sin t - 1}{a \cos^3 t}$.

(D) Given $x=a t^2$ and $y=2 a t$.
First,find $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.
Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{t}) = -\frac{1}{t^2} \cdot \frac{dt}{dx}$.
Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2at}$,we have:
$\frac{d^2y}{dx^2} = -\frac{1}{t^2} \cdot \frac{1}{2at} = -\frac{1}{2at^3}$.
At $t=\frac{1}{2}$,$\frac{d^2y}{dx^2} = -\frac{1}{2a(1/2)^3} = -\frac{1}{2a(1/8)} = -\frac{1}{a/4} = -\frac{4}{a}$.

(C) Given,$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$.
Let $t = \tan \theta$,then $\theta = \tan^{-1} t$.
For $x$,we have $x = \cos^{-1}\left(\frac{1}{\sqrt{1+\tan^2 \theta}}\right) = \cos^{-1}\left(\frac{1}{\sec \theta}\right) = \cos^{-1}(\cos \theta) = \theta = \tan^{-1} t$.
For $y$,we have $y = \sin^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan^2 \theta}}\right) = \sin^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) = \sin^{-1}(\sin \theta) = \theta = \tan^{-1} t$.
Since $x = \tan^{-1} t$ and $y = \tan^{-1} t$,it follows that $y = x$.
Therefore,differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

(D) Given $f(x) = 5 \cos^3 x - 3 \sin^2 x$ and $g(x) = 4 \sin^3 x + \cos^2 x$.
First,find the derivative of $f(x)$ with respect to $x$:
$\frac{df}{dx} = 5(3 \cos^2 x)(-\sin x) - 3(2 \sin x \cos x) = -15 \cos^2 x \sin x - 6 \sin x \cos x$.
Next,find the derivative of $g(x)$ with respect to $x$:
$\frac{dg}{dx} = 4(3 \sin^2 x)(\cos x) + 2 \cos x(-\sin x) = 12 \sin^2 x \cos x - 2 \sin x \cos x$.
Now,find the derivative of $f(x)$ with respect to $g(x)$ using the chain rule:
$\frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{-15 \cos^2 x \sin x - 6 \sin x \cos x}{12 \sin^2 x \cos x - 2 \sin x \cos x}$.
Factor out $-\sin x \cos x$ from the numerator and $\sin x \cos x$ from the denominator:
$\frac{df}{dg} = \frac{-\sin x \cos x (15 \cos x + 6)}{\sin x \cos x (12 \sin x - 2)} = -\frac{15 \cos x + 6}{12 \sin x - 2}$.
Thus,the correct option is $D$.

(B) Given $y = \sin \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)$ and $x = \cos 2 \theta$.
Substitute $x = \cos 2 \theta$ into the expression for $y$:
$\frac{1-x}{1+x} = \frac{1-\cos 2 \theta}{1+\cos 2 \theta} = \frac{2 \sin ^2 \theta}{2 \cos ^2 \theta} = \tan ^2 \theta$.
Thus,$\tan ^{-1} \sqrt{\frac{1-x}{1+x}} = \tan ^{-1}(\tan \theta) = \theta$.
Therefore,$y = \sin (2 \theta)$.
Now,differentiate $y$ and $x$ with respect to $\theta$:
$\frac{d y}{d \theta} = \frac{d}{d \theta} (\sin 2 \theta) = 2 \cos 2 \theta$.
$\frac{d x}{d \theta} = \frac{d}{d \theta} (\cos 2 \theta) = -2 \sin 2 \theta$.
Using the chain rule,$\frac{d y}{d x} = \frac{d y / d \theta}{d x / d \theta} = \frac{2 \cos 2 \theta}{-2 \sin 2 \theta} = -\cot 2 \theta$.

(B) Given $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$.
We know that $\sec \theta = \frac{1}{\cos \theta}$,so $x = \frac{1}{\cos \theta} - \cos \theta = \frac{\sin^2 \theta}{\cos \theta}$.
Then $\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \frac{\sin \theta}{\cos^2 \theta} + \sin \theta = \sin \theta \left(\frac{1 + \cos^2 \theta}{\cos^2 \theta}\right)$.
Similarly,$\frac{dy}{d\theta} = n \sec^n \theta \tan \theta + n \cos^{n-1} \theta \sin \theta = n \sin \theta \left(\frac{1 + \cos^{2n} \theta}{\cos^{n+1} \theta}\right)$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{n \sin \theta (1 + \cos^{2n} \theta) / \cos^{n+1} \theta}{\sin \theta (1 + \cos^2 \theta) / \cos^2 \theta} = \frac{n (1 + \cos^{2n} \theta)}{\cos^{n-1} \theta (1 + \cos^2 \theta)}$.
Also,$x^2 + 4 = (\sec \theta - \cos \theta)^2 + 4 = \sec^2 \theta - 2 + \cos^2 \theta + 4 = \sec^2 \theta + 2 + \cos^2 \theta = (\sec \theta + \cos \theta)^2 = \left(\frac{1 + \cos^2 \theta}{\cos \theta}\right)^2$.
Thus,$(x^2 + 4) \left(\frac{dy}{dx}\right)^2 = \left(\frac{1 + \cos^2 \theta}{\cos \theta}\right)^2 \cdot \frac{n^2 (1 + \cos^{2n} \theta)^2}{\cos^{2n-2} \theta (1 + \cos^2 \theta)^2} = \frac{n^2 (1 + \cos^{2n} \theta)^2}{\cos^{2n} \theta} = n^2 \left(\frac{1}{\cos^n \theta} - \cos^n \theta\right)^2 + 4n^2 = n^2 (y^2 + 4)$.

(C) Given: $y=4 \cos ^3(t)$ and $x=4 \sin ^3(t)$.
First,differentiate $y$ with respect to $t$:
$\frac{d y}{d t} = 4 \cdot 3 \cos ^2(t) \cdot (-\sin(t)) = -12 \sin(t) \cos ^2(t)$.
Next,differentiate $x$ with respect to $t$:
$\frac{d x}{d t} = 4 \cdot 3 \sin ^2(t) \cdot \cos(t) = 12 \sin ^2(t) \cos(t)$.
Using the chain rule for parametric differentiation:
$\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{-12 \sin(t) \cos ^2(t)}{12 \sin ^2(t) \cos(t)}$.
Simplifying the expression:
$\frac{d y}{d x} = -\frac{\cos(t)}{\sin(t)} = -\cot(t)$.
Thus,the correct option is $C$.

(D) Let $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ and $v = \sqrt{1-x^2}$.
Substitute $x = \cos \theta$.
Then $u = \sec ^{-1}\left(\frac{1}{2 \cos ^2 \theta - 1}\right) = \sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right) = \sec ^{-1}(\sec 2 \theta) = 2 \theta$.
And $v = \sqrt{1 - \cos ^2 \theta} = \sqrt{\sin ^2 \theta} = \sin \theta$.
Now,differentiate $u$ and $v$ with respect to $\theta$:
$\frac{du}{d\theta} = 2$ and $\frac{dv}{d\theta} = \cos \theta$.
Therefore,the derivative of $u$ with respect to $v$ is:
$\frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{2}{\cos \theta} = \frac{2}{x}$.
At $x = \frac{1}{2}$,we have:
$\left(\frac{du}{dv}\right)_{x=1/2} = \frac{2}{1/2} = 4$.

(A) Given equations are:
$x = a(t + \sin t) \quad \dots (i)$
$y = a(1 - \cos t) \quad \dots (ii)$
Differentiating $(i)$ with respect to $t$:
$\frac{dx}{dt} = a(1 + \cos t)$
Differentiating $(ii)$ with respect to $t$:
$\frac{dy}{dt} = a(\sin t)$
Now,find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t} = \tan\left(\frac{t}{2}\right)$
Now,find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\tan\left(\frac{t}{2}\right)\right) = \frac{d}{dt}\left(\tan\left(\frac{t}{2}\right)\right) \cdot \frac{dt}{dx}$
$= \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2} \cdot \frac{1}{a(1 + \cos t)}$
$= \frac{1}{2 \cos^2(t/2)} \cdot \frac{1}{a(2 \cos^2(t/2))} = \frac{1}{4a \cos^4(t/2)}$
At $t = \frac{2\pi}{3}$,$\frac{t}{2} = \frac{\pi}{3}$:
$\frac{d^2y}{dx^2} = \frac{1}{4a \cos^4(\pi/3)} = \frac{1}{4a (1/2)^4} = \frac{1}{4a (1/16)} = \frac{16}{4a} = \frac{4}{a}$

(C) Given,$\cos (f(x)) = \frac{1-x^2}{1+x^2}$.
Let $x = \tan \theta$,then $\cos (f(x)) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta$.
Thus,$f(x) = 2\theta = 2 \tan^{-1} x$.
Now,$\tan (g(x)) = \frac{3x-x^3}{1-3x^2}$.
Using the identity $\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}$,we get $\tan (g(x)) = \tan 3\theta$.
Thus,$g(x) = 3\theta = 3 \tan^{-1} x$.
We need to find $\frac{df}{dg} = \frac{df/dx}{dg/dx}$.
$\frac{df}{dx} = \frac{d}{dx}(2 \tan^{-1} x) = \frac{2}{1+x^2}$.
$\frac{dg}{dx} = \frac{d}{dx}(3 \tan^{-1} x) = \frac{3}{1+x^2}$.
Therefore,$\frac{df}{dg} = \frac{2/(1+x^2)}{3/(1+x^2)} = \frac{2}{3}$.

(B) Given $x=a(t+\sin t)$ and $y=a(1-\cos t)$.
First,find the derivatives with respect to $t$:
$\frac{d x}{d t}=a(1+\cos t)$ and $\frac{d y}{d t}=a \sin t$.
Now,find $\frac{d y}{d x}$:
$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)}=\tan(t/2)$.
Next,differentiate $\frac{d y}{d x}$ with respect to $x$:
$\frac{d^2 y}{d x^2}=\frac{d}{d x}(\tan(t/2))=\sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{d t}{d x}$.
Since $\frac{d x}{d t}=a(1+\cos t)=2a \cos^2(t/2)$,then $\frac{d t}{d x}=\frac{1}{2a \cos^2(t/2)}$.
Substituting this back:
$\frac{d^2 y}{d x^2}=\sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{1}{2a \cos^2(t/2)}=\frac{1}{4a \cos^4(t/2)}$.

(C) Given $x = \sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$.
Squaring both sides,we get $x^2 = 2^{\operatorname{cosec}^{-1} t}$ and $y^2 = 2^{\sec^{-1} t}$.
Multiplying the two equations: $x^2 y^2 = 2^{\operatorname{cosec}^{-1} t + \sec^{-1} t}$.
Since $\operatorname{cosec}^{-1} t + \sec^{-1} t = \frac{\pi}{2}$ for $|t| \geq 1$,we have $x^2 y^2 = 2^{\pi/2}$.
Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(2^{\pi/2})$.
$x^2 (2y \frac{dy}{dx}) + y^2 (2x) = 0$.
$2x^2 y \frac{dy}{dx} = -2xy^2$.
$\frac{dy}{dx} = -\frac{2xy^2}{2x^2 y} = -\frac{y}{x}$.

(B) Given,$x=4 \cos ^3 \theta$ ... $(i)$ and $y=3 \sin ^2 \theta$ ... (ii).
Differentiating $(i)$ and (ii) with respect to $\theta$:
$\frac{dx}{d\theta} = 4 \times 3 \cos^2 \theta (-\sin \theta) = -12 \cos^2 \theta \sin \theta$ ... (iii).
$\frac{dy}{d\theta} = 3 \times 2 \sin \theta \cos \theta = 6 \sin \theta \cos \theta$ ... (iv).
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{6 \sin \theta \cos \theta}{-12 \cos^2 \theta \sin \theta} = -\frac{1}{2} \sec \theta$.
Differentiating $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{1}{2} \sec \theta \right) = -\frac{1}{2} \sec \theta \tan \theta \cdot \frac{d\theta}{dx}$.
Substituting $\frac{d\theta}{dx} = \frac{1}{-12 \cos^2 \theta \sin \theta}$:
$\frac{d^2y}{dx^2} = -\frac{1}{2} \sec \theta \tan \theta \cdot \left( \frac{1}{-12 \cos^2 \theta \sin \theta} \right) = \frac{1}{24} \cdot \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos^2 \theta \sin \theta} = \frac{1}{24 \cos^4 \theta}$.
At $\theta = \frac{\pi}{4}$,$\cos \theta = \frac{1}{\sqrt{2}}$,so $\cos^4 \theta = \left( \frac{1}{\sqrt{2}} \right)^4 = \frac{1}{4}$.
Thus,$\frac{d^2y}{dx^2} = \frac{1}{24 \times (1/4)} = \frac{1}{6}$.

(D) Given,$x=\cos \theta$ and $y=\sin 5 \theta$.
Differentiating with respect to $\theta$:
$\frac{d x}{d \theta}=-\sin \theta$ and $\frac{d y}{d \theta}=5 \cos 5 \theta$.
Therefore,$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta} = \frac{5 \cos 5 \theta}{-\sin \theta} = -5 \cos 5 \theta \csc \theta$.
Now,find $\frac{d^2 y}{d x^2} = \frac{d}{d \theta} \left( \frac{d y}{d x} \right) \cdot \frac{d \theta}{d x}$.
$\frac{d}{d \theta} (-5 \cos 5 \theta \csc \theta) = -5 [(-5 \sin 5 \theta) \csc \theta + \cos 5 \theta (-\csc \theta \cot \theta)] = 25 \sin 5 \theta \csc \theta + 5 \cos 5 \theta \csc \theta \cot \theta$.
Multiplying by $\frac{d \theta}{d x} = \frac{1}{-\sin \theta} = -\csc \theta$:
$\frac{d^2 y}{d x^2} = -25 \sin 5 \theta \csc^2 \theta - 5 \cos 5 \theta \csc^2 \theta \cot \theta$.
Substitute into the expression $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$:
Since $1-x^2 = \sin^2 \theta$ and $x = \cos \theta$:
$\sin^2 \theta (-25 \sin 5 \theta \csc^2 \theta - 5 \cos 5 \theta \csc^2 \theta \cot \theta) - \cos \theta (-5 \cos 5 \theta \csc \theta)$
$= -25 \sin 5 \theta - 5 \cos 5 \theta \cot \theta + 5 \cos 5 \theta \cot \theta$
$= -25 \sin 5 \theta = -25 y$.

(D) Given the parametric equations of the curve: $x=e^t \cos t$ and $y=e^t \sin t$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = e^t(\cos t - \sin t)$ and $\frac{dy}{dt} = e^t(\sin t + \cos t)$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin t + \cos t}{\cos t - \sin t}$.
At the point $(1,0)$,we have $y=0$,which implies $e^t \sin t = 0$. Since $e^t \neq 0$,we have $\sin t = 0$,so $t=0$.
At $t=0$,the slope of the tangent is $\frac{dy}{dx} = \frac{\sin 0 + \cos 0}{\cos 0 - \sin 0} = \frac{0+1}{1-0} = 1$.
The slope of the normal $m$ is given by $m = -\frac{1}{dy/dx} = -\frac{1}{1} = -1$.
Since the slope of the normal is $\tan \theta = -1$,and $\theta$ is the angle with the $X$-axis,we have $\theta = \frac{3\pi}{4}$.

(A) Given the parametric equations of the curve:
$x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a(1 + \cos \theta)$
$\frac{dy}{d\theta} = a \sin \theta$
Now,the slope of the tangent is given by:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}$
Using the trigonometric identity $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 + \cos \theta = 2 \cos^2(\theta/2)$:
$\frac{dy}{dx} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)} = \tan(\theta/2)$
At $\theta = \pi / 3$:
$\frac{dy}{dx} = \tan(\frac{\pi/3}{2}) = \tan(\pi / 6)$
Since the slope $m = \tan \psi$,where $\psi$ is the angle with the $X$-axis:
$\tan \psi = \tan(\pi / 6) \Rightarrow \psi = \pi / 6$.

(B) Given $x=5(1-\sin t)$ and $y=5(t+\cos t)$.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = 5(0 - \cos t) = -5\cos t$ ... $(i)$
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = 5(1 - \sin t)$ ... $(ii)$
Using the chain rule,we find $\frac{dx}{dy}$:
$\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{-5\cos t}{5(1-\sin t)}$
$\frac{dx}{dy} = \frac{-\cos t}{1-\sin t} = \frac{\cos t}{\sin t-1}$

(B) Given $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$.
Multiplying both sides by $(1 + \sqrt[3]{y})$,we get $x(1 + \sqrt[3]{y}) = 1 - \sqrt[3]{y}$.
$x + x\sqrt[3]{y} = 1 - \sqrt[3]{y}$.
$\sqrt[3]{y}(x + 1) = 1 - x$.
$\sqrt[3]{y} = \frac{1 - x}{1 + x}$.
Cubing both sides,$y = \left(\frac{1 - x}{1 + x}\right)^3$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 3\left(\frac{1 - x}{1 + x}\right)^2 \cdot \frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)$.
Using the quotient rule,$\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right) = \frac{-(1 + x) - (1 - x)(1)}{(1 + x)^2} = \frac{-1 - x - 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2}$.
Thus,$\frac{dy}{dx} = 3 \cdot \frac{(1 - x)^2}{(1 + x)^2} \cdot \frac{-2}{(1 + x)^2} = \frac{-6(1 - x)^2}{(1 + x)^4}$.
Since $\frac{dx}{dy} = \frac{1}{dy/dx}$,we have $\frac{dx}{dy} = \frac{-(1 + x)^4}{6(1 - x)^2}$.

(B) Given $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$.
Using the quotient rule $\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$\frac{dx}{dt} = \frac{(1+t^5)(2t) - t^2(5t^4)}{(1+t^5)^2} = \frac{2t + 2t^6 - 5t^6}{(1+t^5)^2} = \frac{2t - 3t^6}{(1+t^5)^2} = \frac{t(2 - 3t^5)}{(1+t^5)^2}$.
$\frac{dy}{dt} = \frac{(1+t^5)(6t^2) - 2t^3(5t^4)}{(1+t^5)^2} = \frac{6t^2 + 6t^7 - 10t^7}{(1+t^5)^2} = \frac{6t^2 - 4t^7}{(1+t^5)^2} = \frac{2t^2(3 - 2t^5)}{(1+t^5)^2}$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t^2(3 - 2t^5)}{(1+t^5)^2} \times \frac{(1+t^5)^2}{t(2 - 3t^5)} = \frac{2t(3 - 2t^5)}{(2 - 3t^5)}$.

(B) Given $x = \sin 2\theta \cos 3\theta$ and $y = \sin 3\theta \cos 2\theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = \frac{d}{d\theta}(\sin 2\theta \cos 3\theta) = \cos 2\theta(2) \cos 3\theta + \sin 2\theta(-\sin 3\theta)(3) = 2\cos 2\theta \cos 3\theta - 3\sin 2\theta \sin 3\theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = \frac{d}{d\theta}(\sin 3\theta \cos 2\theta) = \cos 3\theta(3) \cos 2\theta + \sin 3\theta(-\sin 2\theta)(2) = 3\cos 3\theta \cos 2\theta - 2\sin 3\theta \sin 2\theta$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3\cos 3\theta \cos 2\theta - 2\sin 3\theta \sin 2\theta}{2\cos 2\theta \cos 3\theta - 3\sin 2\theta \sin 3\theta}$.
Thus,the correct option is $B$.

(B) Given $x = 2 \sqrt{2} \sqrt{\cos 2 \theta}$ and $y = 2 \sqrt{2} \sqrt{\sin 2 \theta}$.
Squaring both equations,we get $x^2 = 8 \cos 2 \theta$ and $y^2 = 8 \sin 2 \theta$.
Adding the squares: $x^2 + y^2 = 8(\cos 2 \theta + \sin 2 \theta)$.
Alternatively,differentiate with respect to $\theta$:
$\frac{dx}{d \theta} = 2 \sqrt{2} \cdot \frac{1}{2 \sqrt{\cos 2 \theta}} \cdot (-2 \sin 2 \theta) = -\frac{2 \sqrt{2} \sin 2 \theta}{\sqrt{\cos 2 \theta}}$.
$\frac{dy}{d \theta} = 2 \sqrt{2} \cdot \frac{1}{2 \sqrt{\sin 2 \theta}} \cdot (2 \cos 2 \theta) = \frac{2 \sqrt{2} \cos 2 \theta}{\sqrt{\sin 2 \theta}}$.
Then $\frac{dy}{dx} = \frac{dy/d \theta}{dx/d \theta} = \frac{2 \sqrt{2} \cos 2 \theta / \sqrt{\sin 2 \theta}}{-2 \sqrt{2} \sin 2 \theta / \sqrt{\cos 2 \theta}} = -\frac{\cos 2 \theta \sqrt{\cos 2 \theta}}{\sin 2 \theta \sqrt{\sin 2 \theta}} = -\cot 2 \theta \sqrt{\cot 2 \theta} = -(\cot 2 \theta)^{3/2}$.
At $\theta = 22 \frac{1}{2}^{\circ} = \frac{45^{\circ}}{2}$,$2 \theta = 45^{\circ}$.
So,$\frac{dy}{dx} = -(\cot 45^{\circ})^{3/2} = -(1)^{3/2} = -1$.

(D) Given $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$.
First,differentiate $y$ with respect to $t$ using the quotient rule:
$\frac{dy}{dt} = \frac{(1-t^4)(32t) - (16t^2)(-4t^3)}{(1-t^4)^2} = \frac{32t - 32t^5 + 64t^5}{(1-t^4)^2} = \frac{32t(1+t^4)}{(1-t^4)^2}$.
Next,differentiate $x$ with respect to $t$ using the quotient rule:
$\frac{dx}{dt} = \frac{(1+t^4)(18t) - (9t^2)(4t^3)}{(1+t^4)^2} = \frac{18t + 18t^5 - 36t^5}{(1+t^4)^2} = \frac{18t(1-t^4)}{(1+t^4)^2}$.
Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{32t(1+t^4)}{(1-t^4)^2} \times \frac{(1+t^4)^2}{18t(1-t^4)} = \frac{32(1+t^4)^3}{18(1-t^4)^3} = \frac{16}{9}\left(\frac{1+t^4}{1-t^4}\right)^3$.

(B) Given $x = \cos 2t + \log(\tan t)$ and $y = 2t + \cot 2t$.
First,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(2t + \cot 2t) = 2 - 2\operatorname{cosec}^2 2t = -2(\operatorname{cosec}^2 2t - 1) = -2\cot^2 2t$.
Next,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(\cos 2t + \log(\tan t)) = -2\sin 2t + \frac{1}{\tan t} \cdot \sec^2 t = -2\sin 2t + \frac{\cos t}{\sin t} \cdot \frac{1}{\cos^2 t} = -2\sin 2t + \frac{1}{\sin t \cos t}$.
Since $\sin 2t = 2\sin t \cos t$,we have $\frac{1}{\sin t \cos t} = \frac{2}{\sin 2t} = 2\operatorname{cosec} 2t$.
So,$\frac{dx}{dt} = -2\sin 2t + 2\operatorname{cosec} 2t = 2(\operatorname{cosec} 2t - \sin 2t) = 2\left(\frac{1 - \sin^2 2t}{\sin 2t}\right) = 2\frac{\cos^2 2t}{\sin 2t} = 2\cot 2t \cos 2t$.
Finally,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\cot^2 2t}{2\cot 2t \cos 2t} = -\frac{\cot 2t}{\cos 2t} = -\frac{\cos 2t}{\sin 2t} \cdot \frac{1}{\cos 2t} = -\frac{1}{\sin 2t} = -\operatorname{cosec} 2t$.

(C) Given $x = 3 \sqrt{2} \cos^3 \theta$ and $y = 4 \tan^2 \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = 3 \sqrt{2} \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -9 \sqrt{2} \cos^2 \theta \sin \theta$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = 4 \cdot 2 \tan \theta \cdot \sec^2 \theta = 8 \tan \theta \sec^2 \theta$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{8 \tan \theta \sec^2 \theta}{-9 \sqrt{2} \cos^2 \theta \sin \theta}$.
At $\theta = \frac{\pi}{4}$,we have $\tan \frac{\pi}{4} = 1$,$\sec \frac{\pi}{4} = \sqrt{2}$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
Substituting these values:
$\left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{4}} = \frac{8(1)(\sqrt{2})^2}{-9 \sqrt{2} (\frac{1}{\sqrt{2}})^2 (\frac{1}{\sqrt{2}})} = \frac{8 \cdot 2}{-9 \sqrt{2} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}} = \frac{16}{-9 \cdot \frac{1}{2}} = \frac{16}{-4.5} = -\frac{32}{9}$.

(A) Given that $x = \log p$ and $y = \frac{1}{p}$.
First,differentiate $y$ with respect to $p$:
$\frac{dy}{dp} = \frac{d}{dp}(p^{-1}) = -p^{-2} = -\frac{1}{p^2}$.
Next,differentiate $x$ with respect to $p$:
$\frac{dx}{dp} = \frac{d}{dp}(\log p) = \frac{1}{p}$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/dp}{dx/dp} = \frac{-1/p^2}{1/p} = -\frac{1}{p}$.
Since $x = \log p$,we have $p = e^x$.
Substituting $p = e^x$ into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{1}{e^x} = -e^{-x}$.