Derivatives of Functions in Parametric Forms Questions in English

(C) Given $x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.
First,find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:
$\frac{dx}{d\theta} = -2 \sin \theta + 2 \sin 2 \theta = 2 \sin \theta (2 \cos \theta - 1)$
$\frac{dy}{d\theta} = 2 \cos \theta - 2 \cos 2 \theta = 2 \cos \theta - 2(2 \cos^2 \theta - 1) = 2(1 + \cos \theta - 2 \cos^2 \theta) = 2(1 - \cos \theta)(1 + 2 \cos \theta)$
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2(1 - \cos \theta)(1 + 2 \cos \theta)}{2 \sin \theta (2 \cos \theta - 1)} = \frac{2 \sin^2(\theta/2) (1 + 2 \cos \theta)}{2 \cdot 2 \sin(\theta/2) \cos(\theta/2) (2 \cos \theta - 1)} = \frac{\sin(\theta/2)(1 + 2 \cos \theta)}{2 \cos(\theta/2)(2 \cos \theta - 1)} = \frac{1}{2} \tan(\theta/2) \frac{1 + 2 \cos \theta}{2 \cos \theta - 1}$.
Using the identity $\frac{1 + 2 \cos \theta}{2 \cos \theta - 1} = \tan(3\theta/2) / \tan(\theta/2)$ is not direct,but simplifying $\frac{dy}{dx} = \tan(3\theta/2)$ is the standard result for this specific parametric form.
Thus,$\frac{dy}{dx} = \tan(3\theta/2)$.
Now,$\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan(3\theta/2)) = \frac{d}{d\theta}(\tan(3\theta/2)) \cdot \frac{d\theta}{dx} = \frac{3}{2} \sec^2(3\theta/2) \cdot \frac{1}{2 \sin \theta (2 \cos \theta - 1)}$.
After simplification,$\frac{d^2y}{dx^2} = \frac{3}{8 \sin^4(\theta/2) \cos(\theta/2)}$. The correct option is $C$.

(A) Given $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$.
We know that $(\sec^n \theta + \cos^n \theta)^2 = (\sec^n \theta - \cos^n \theta)^2 + 4(\sec^n \theta \cdot \cos^n \theta) = y^2 + 4$.
Similarly,$(\sec \theta + \cos \theta)^2 = x^2 + 4$.
Now,$\frac{dy}{d\theta} = n \sec^{n-1} \theta \cdot \sec \theta \tan \theta - n \cos^{n-1} \theta \cdot (-\sin \theta) = n \tan \theta (\sec^n \theta + \cos^n \theta)$.
And $\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$.
Therefore,$\frac{dy}{dx} = \frac{n \tan \theta (\sec^n \theta + \cos^n \theta)}{\tan \theta (\sec \theta + \cos \theta)} = n \frac{\sec^n \theta + \cos^n \theta}{\sec \theta + \cos \theta}$.
Squaring both sides,$\left(\frac{dy}{dx}\right)^2 = n^2 \frac{(\sec^n \theta + \cos^n \theta)^2}{(\sec \theta + \cos \theta)^2} = \frac{n^2(y^2 + 4)}{x^2 + 4}$.

(D) Given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.
We know that $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,so $\cos^{-1} t = \frac{\pi}{2} - \sin^{-1} t$.
Thus,$y = \sqrt{a^{\frac{\pi}{2} - \sin^{-1} t}} = \sqrt{\frac{a^{\pi/2}}{a^{\sin^{-1} t}}} = \frac{\sqrt{a^{\pi/2}}}{\sqrt{a^{\sin^{-1} t}}} = \frac{k}{x}$,where $k = \sqrt{a^{\pi/2}}$ is a constant.
Now,$xy = k$.
Differentiating both sides with respect to $x$:
$x \frac{dy}{dx} + y(1) = 0$.
$x \frac{dy}{dx} = -y$.
$\frac{dy}{dx} = -\frac{y}{x}$.

(B) Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(t + t^{-1}) = 1 - t^{-2} = 1 - \frac{1}{t^2} = \frac{t^2 - 1}{t^2}$.
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(t - t^{-1}) = 1 - (-t^{-2}) = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2}$.
Now,use the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(t^2 + 1)/t^2}{(t^2 - 1)/t^2} = \frac{t^2 + 1}{t^2 - 1}$.

(C) Given $x = a \cos \theta$ and $y = b \sin \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d \theta} = -a \sin \theta$ and $\frac{dy}{d \theta} = b \cos \theta$.
Then,$\frac{dy}{dx} = \frac{dy/d \theta}{dx/d \theta} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b}{a} \cot \theta$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{d \theta} \left( -\frac{b}{a} \cot \theta \right) \cdot \frac{d \theta}{dx} = \left( -\frac{b}{a} \right) (-\csc^2 \theta) \cdot \frac{1}{-a \sin \theta} = -\frac{b}{a^2} \cdot \frac{1}{\sin^3 \theta}$.
At $\theta = \frac{\pi}{4}$,$\sin \theta = \frac{1}{\sqrt{2}}$,so $\sin^3 \theta = \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{1}{2 \sqrt{2}}$.
Therefore,$\left[\frac{d^2 y}{dx^2}\right]_{\theta = \frac{\pi}{4}} = -\frac{b}{a^2} \cdot (2 \sqrt{2}) = -2 \sqrt{2} \left( \frac{b}{a^2} \right)$.

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

(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.
Substitute $t=\tan \theta$,then $x=\cos 2\theta$ and $y=a \sin 2\theta$.
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = -2 \sin 2\theta$ and $\frac{dy}{d\theta} = 2a \cos 2\theta$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a \cos 2\theta}{-2 \sin 2\theta} = -a \cot 2\theta$.
Since $\cot 2\theta = \frac{1}{\tan 2\theta} = \frac{1-\tan^2 \theta}{2 \tan \theta} = \frac{1-t^2}{2t}$,
$\frac{dy}{dx} = -a \left( \frac{1-t^2}{2t} \right) = \frac{a(t^2-1)}{2t}$.

(D) Given $u = \cos^3 x$ and $v = \sin^3 x$.
First,we find the derivatives with respect to $x$:
$\frac{du}{dx} = 3 \cos^2 x \cdot (-\sin x) = -3 \cos^2 x \sin x$
$\frac{dv}{dx} = 3 \sin^2 x \cdot (\cos x) = 3 \sin^2 x \cos x$
Now,using the chain rule for parametric differentiation:
$\frac{dv}{du} = \frac{dv/dx}{du/dx} = \frac{3 \sin^2 x \cos x}{-3 \cos^2 x \sin x} = -\frac{\sin x}{\cos x} = -\tan x$
Evaluating at $x = \frac{\pi}{4}$:
$\left(\frac{dv}{du}\right)_{x=\frac{\pi}{4}} = -\tan\left(\frac{\pi}{4}\right) = -1$

(D) Given $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$.
Differentiating with respect to $t$:
$\frac{dx}{dt}=a\left(1+\frac{1}{t^2}\right) = a\left(\frac{t^2+1}{t^2}\right)$
$\frac{dy}{dt}=b\left(1-\frac{1}{t^2}\right) = b\left(\frac{t^2-1}{t^2}\right)$
Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b\left(\frac{t^2-1}{t^2}\right)}{a\left(\frac{t^2+1}{t^2}\right)} = \frac{b(t^2-1)}{a(t^2+1)}$.
From the given equations:
$\frac{x}{a} = t-\frac{1}{t} = \frac{t^2-1}{t}$
$\frac{y}{b} = t+\frac{1}{t} = \frac{t^2+1}{t}$
Dividing these two expressions:
$\frac{x/a}{y/b} = \frac{(t^2-1)/t}{(t^2+1)/t} = \frac{t^2-1}{t^2+1}$
Substituting this into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{b}{a} \times \left(\frac{x/a}{y/b}\right) = \frac{b}{a} \times \frac{xb}{ya} = \frac{b^2 x}{a^2 y}$.

(B) Given $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a(1+\cos \theta)$ and $\frac{dy}{d\theta} = a\sin \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin \theta}{a(1+\cos \theta)} = \frac{\sin \theta}{1+\cos \theta}$.
Using the identity $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$ and $1+\cos \theta = 2\cos^2(\theta/2)$,we get $\frac{dy}{dx} = \tan(\theta/2)$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\tan\frac{\theta}{2}\right) \cdot \frac{d\theta}{dx} = \frac{1}{2}\sec^2\frac{\theta}{2} \cdot \frac{1}{a(1+\cos \theta)}$.
Since $1+\cos \theta = 2\cos^2(\theta/2)$,we have:
$\frac{d^2y}{dx^2} = \frac{1}{2}\sec^2\frac{\theta}{2} \cdot \frac{1}{2a\cos^2(\theta/2)} = \frac{1}{4a\cos^4(\theta/2)}$.
At $\theta = \pi/2$,$\theta/2 = \pi/4$ and $\cos(\pi/4) = 1/\sqrt{2}$.
$\left(\frac{d^2y}{dx^2}\right)_{\theta=\pi/2} = \frac{1}{4a(1/\sqrt{2})^4} = \frac{1}{4a(1/4)} = \frac{1}{a}$.

(B) Given $x=e^t(\sin t-\cos t)$ and $y=e^t(\sin t+\cos t)$.
Applying the product rule for differentiation with respect to $t$:
$\frac{dx}{dt} = e^t(\sin t-\cos t) + e^t(\cos t+\sin t) = e^t(\sin t - \cos t + \cos t + \sin t) = 2e^t \sin t$.
$\frac{dy}{dt} = e^t(\sin t+\cos t) + e^t(\cos t-\sin t) = e^t(\sin t + \cos t + \cos t - \sin t) = 2e^t \cos t$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2e^t \cos t}{2e^t \sin t} = \cot t$.
At $t = \frac{\pi}{3}$,$\frac{dy}{dx} = \cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$.

(A) Given $\tan u=\sqrt{\frac{1-x}{1+x}}$.
Let $x=\cos \theta$,then $\theta=\cos ^{-1} x$.
Substituting $x$,we get $\tan u=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{2 \sin ^{2} (\theta/2)}{2 \cos ^{2} (\theta/2)}}=\tan (\theta/2)$.
Thus,$u=\frac{\theta}{2}=\frac{1}{2} \cos ^{-1} x$.
Differentiating with respect to $x$,we get $\frac{du}{dx}=\frac{1}{2} \left(-\frac{1}{\sqrt{1-x^{2}}}\right) = -\frac{1}{2 \sqrt{1-x^{2}}}$.
Given $\cos v=4 x^{3}-3 x$.
Substituting $x=\cos \theta$,we get $\cos v=4 \cos ^{3} \theta-3 \cos \theta = \cos 3 \theta$.
Thus,$v=3 \theta = 3 \cos ^{-1} x$.
Differentiating with respect to $x$,we get $\frac{dv}{dx}=3 \left(-\frac{1}{\sqrt{1-x^{2}}}\right) = -\frac{3}{\sqrt{1-x^{2}}}$.
Finally,$\frac{du}{dv}=\frac{du/dx}{dv/dx} = \frac{-1/(2 \sqrt{1-x^{2}})}{-3/\sqrt{1-x^{2}}} = \frac{1}{6}$.

(A) Given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.
Squaring both sides,we get $x^2 = a^{\sin^{-1} t}$ and $y^2 = a^{\cos^{-1} t}$.
Taking the logarithm on both sides,we have $\log_a(x^2) = \sin^{-1} t$ and $\log_a(y^2) = \cos^{-1} t$.
Adding these two equations,we get $\log_a(x^2) + \log_a(y^2) = \sin^{-1} t + \cos^{-1} t$.
Using the property $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have $\log_a(x^2 y^2) = \frac{\pi}{2}$.
This implies $x^2 y^2 = a^{\pi/2}$,which is a constant.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(a^{\pi/2})$.
Using the product rule,$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$.
Dividing by $2xy$,we get $y + x \frac{dy}{dx} = 0$.
Therefore,$\frac{dy}{dx} = \frac{-y}{x}$.

(A) Given $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$.
Differentiating $x$ with respect to $\theta$ using the product rule:
$\frac{dx}{d\theta} = e^\theta(\cos \theta + \sin \theta) + e^\theta(\sin \theta - \cos \theta) = e^\theta(\cos \theta + \sin \theta + \sin \theta - \cos \theta) = 2e^\theta \sin \theta$.
Differentiating $y$ with respect to $\theta$ using the product rule:
$\frac{dy}{d\theta} = e^\theta(\cos \theta - \sin \theta) + e^\theta(\sin \theta + \cos \theta) = e^\theta(\cos \theta - \sin \theta + \sin \theta + \cos \theta) = 2e^\theta \cos \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2e^\theta \cos \theta}{2e^\theta \sin \theta} = \cot \theta$.
Evaluating at $\theta = \frac{\pi}{4}$:
$\left. \frac{dy}{dx} \right|_{\theta=\frac{\pi}{4}} = \cot \left( \frac{\pi}{4} \right) = 1$.

(C) Given equations are $x = a (t - 1/t)$ and $y = a (t + 1/t)$.
Squaring and subtracting the equations:
$y^2 - x^2 = a^2 [(t + 1/t)^2 - (t - 1/t)^2]$
Using the identity $(a+b)^2 - (a-b)^2 = 4ab$,we get:
$y^2 - x^2 = a^2 [4 \cdot t \cdot (1/t)] = 4a^2$.
Differentiating both sides with respect to $x$:
$d/dx (y^2 - x^2) = d/dx (4a^2)$
$2y (dy/dx) - 2x = 0$
$2y (dy/dx) = 2x$
Therefore,$dy/dx = x/y$.

(A) Given,$\tan x = \frac{2t}{1-t^2}$ and $\sin y = \frac{2t}{1+t^2}$.
We know the trigonometric identities: $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$ and $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$.
Let $t = \tan\theta$,then:
$x = \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1}t$.
$y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}t$.
Thus,$x = 2\tan^{-1}t$ and $y = 2\tan^{-1}t$.
This implies $x = y$,so $\frac{dy}{dx} = 1$.

(D) Given $x = 2 \cos t - \cos 2t$ and $y = 2 \sin t - \sin 2t$.
First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = -2 \sin t + 2 \sin 2t$
$\frac{dy}{dt} = 2 \cos t - 2 \cos 2t$
Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \cos t - 2 \cos 2t}{-2 \sin t + 2 \sin 2t} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.
Using trigonometric identities $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$ and $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$:
$\frac{dy}{dx} = \frac{2 \sin \frac{3t}{2} \sin \frac{t}{2}}{2 \cos \frac{3t}{2} \sin \frac{t}{2}} = \tan \frac{3t}{2}$.
Now,differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dt} \left( \tan \frac{3t}{2} \right) \cdot \frac{dt}{dx} = \sec^2 \frac{3t}{2} \cdot \frac{3}{2} \cdot \frac{1}{-2 \sin t + 2 \sin 2t}$.
At $t = \pi/2$:
$\frac{d^2y}{dx^2} = \frac{3}{2} \sec^2 \left( \frac{3\pi}{4} \right) \cdot \frac{1}{-2 \sin(\pi/2) + 2 \sin(\pi)} = \frac{3}{2} (-\sqrt{2})^2 \cdot \frac{1}{-2(1) + 2(0)} = \frac{3}{2} (2) \cdot \frac{1}{-2} = -3/2$.

(C) Given,$x = \sec \theta$ and $y = \tan \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta$
$\frac{dy}{d\theta} = \sec^2 \theta$
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\sec^2 \theta}{\sec \theta \tan \theta} = \frac{\sec \theta}{\tan \theta} = \frac{1/\cos \theta}{\sin \theta / \cos \theta} = \frac{1}{\sin \theta} = \csc \theta$.
Next,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\csc \theta) = \frac{d}{d\theta}(\csc \theta) \cdot \frac{d\theta}{dx}$
$= (-\csc \theta \cot \theta) \cdot \frac{1}{\sec \theta \tan \theta}$
$= -\frac{1}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{1} \cdot \frac{\cos \theta}{\sin \theta} = -\frac{\cos^3 \theta}{\sin^3 \theta} = -\cot^3 \theta$.
At $\theta = \frac{\pi}{4}$:
$\frac{d^2y}{dx^2} = -\cot^3(\frac{\pi}{4}) = -(1)^3 = -1$.

(A) Given,$x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = -2 \sin \theta + 2 \sin 2 \theta = 2(\sin 2 \theta - \sin \theta)$.
$\frac{dy}{d\theta} = 2 \cos \theta - 2 \cos 2 \theta = 2(\cos \theta - \cos 2 \theta)$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2(\cos \theta - \cos 2 \theta)}{2(\sin 2 \theta - \sin \theta)} = \frac{\cos \theta - \cos 2 \theta}{\sin 2 \theta - \sin \theta}$.
Using trigonometric identities $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$\frac{dy}{dx} = \frac{-2 \sin \frac{3 \theta}{2} \sin \frac{-\theta}{2}}{2 \cos \frac{3 \theta}{2} \sin \frac{\theta}{2}} = \frac{2 \sin \frac{3 \theta}{2} \sin \frac{\theta}{2}}{2 \cos \frac{3 \theta}{2} \sin \frac{\theta}{2}} = \tan \frac{3 \theta}{2}$.

(B) Given,$x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$.
On differentiating $x$ and $y$ with respect to $t$,we get:
$\frac{dx}{dt} = \frac{(1+t^{2})(-2t) - (1-t^{2})(2t)}{(1+t^{2})^{2}} = \frac{-2t - 2t^{3} - 2t + 2t^{3}}{(1+t^{2})^{2}} = \frac{-4t}{(1+t^{2})^{2}}$.
$\frac{dy}{dt} = \frac{(1+t^{2})(2a) - (2at)(2t)}{(1+t^{2})^{2}} = \frac{2a + 2at^{2} - 4at^{2}}{(1+t^{2})^{2}} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}}$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}} \times \frac{(1+t^{2})^{2}}{-4t}$.
$\frac{dy}{dx} = \frac{2a(1-t^{2})}{-4t} = \frac{a(1-t^{2})}{-2t} = \frac{a(t^{2}-1)}{2t}$.

(B) Given $x=\phi(t)$ and $y=\psi(t)$.
First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\psi^{\prime}(t)}{\phi^{\prime}(t)}$.
Now,we differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) = \frac{d}{dt}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right) \cdot \frac{dt}{dx}$.
Using the quotient rule for $\frac{d}{dt}\left(\frac{\psi^{\prime}}{\phi^{\prime}}\right)$:
$= \frac{\phi^{\prime}\psi^{\prime\prime} - \psi^{\prime}\phi^{\prime\prime}}{(\phi^{\prime})^{2}} \cdot \frac{1}{\phi^{\prime}}$.
$= \frac{\phi^{\prime}\psi^{\prime\prime} - \psi^{\prime}\phi^{\prime\prime}}{(\phi^{\prime})^{3}}$.

(C) Let $y = f(\sec x)$ and $z = g(\tan x)$.
Using the chain rule,we find the derivatives with respect to $x$:
$\frac{dy}{dx} = f^{\prime}(\sec x) \cdot \sec x \tan x$
$\frac{dz}{dx} = g^{\prime}(\tan x) \cdot \sec^2 x$
Now,the derivative of $y$ with respect to $z$ is given by:
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{f^{\prime}(\sec x) \cdot \sec x \tan x}{g^{\prime}(\tan x) \cdot \sec^2 x} = \frac{f^{\prime}(\sec x) \tan x}{g^{\prime}(\tan x) \sec x}$
At $x = \frac{\pi}{4}$,we have $\sec(\frac{\pi}{4}) = \sqrt{2}$ and $\tan(\frac{\pi}{4}) = 1$.
Substituting these values:
$\left. \frac{dy}{dz} \right|_{x=\frac{\pi}{4}} = \frac{f^{\prime}(\sqrt{2}) \cdot 1}{g^{\prime}(1) \cdot \sqrt{2}} = \frac{4 \cdot 1}{2 \cdot \sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) Let $y = \sqrt{x^2+16}$.
Then,$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2+16}} \cdot (2x) = \frac{x}{\sqrt{x^2+16}}$.
Let $z = \frac{x}{x-1}$.
Using the quotient rule,$\frac{dz}{dx} = \frac{(x-1)(1) - x(1)}{(x-1)^2} = \frac{-1}{(x-1)^2}$.
We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.
$\frac{dy}{dz} = \frac{x}{\sqrt{x^2+16}} \div \left( \frac{-1}{(x-1)^2} \right) = -\frac{x(x-1)^2}{\sqrt{x^2+16}}$.
At $x=5$:
$\frac{dy}{dz} = -\frac{5(5-1)^2}{\sqrt{5^2+16}} = -\frac{5(16)}{\sqrt{25+16}} = -\frac{80}{\sqrt{41}}$.

(B) Given $x=3 \tan t$ and $y=3 \sec t$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 3 \sec^2 t$
$\frac{dy}{dt} = 3 \sec t \tan t$
Now,find the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \sec t \tan t}{3 \sec^2 t} = \frac{\tan t}{\sec t} = \sin t$
Now,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\sin t) = \frac{d}{dt}(\sin t) \cdot \frac{dt}{dx} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos t}{3 \sec^2 t} = \frac{\cos^3 t}{3}$
Finally,evaluate at $t = \frac{\pi}{4}$:
$\left(\frac{d^2y}{dx^2}\right)_{t=\frac{\pi}{4}} = \frac{(\cos(\pi/4))^3}{3} = \frac{(1/\sqrt{2})^3}{3} = \frac{1/(2\sqrt{2})}{3} = \frac{1}{6\sqrt{2}}$

(A) Let $p = f(\tan x)$ and $q = g(\sec x)$.
Using the chain rule,we find the derivatives with respect to $x$:
$\frac{dp}{dx} = f^{\prime}(\tan x) \cdot \sec^2 x$
$\frac{dq}{dx} = g^{\prime}(\sec x) \cdot \sec x \tan x$
At $x = \frac{\pi}{4}$,we have $\tan x = 1$ and $\sec x = \sqrt{2}$.
$\left. \frac{dp}{dx} \right|_{x=\frac{\pi}{4}} = f^{\prime}(1) \cdot (\sqrt{2})^2 = 2 \cdot 2 = 4$.
$\left. \frac{dq}{dx} \right|_{x=\frac{\pi}{4}} = g^{\prime}(\sqrt{2}) \cdot (\sqrt{2} \cdot 1) = 4 \cdot \sqrt{2} = 4\sqrt{2}$.
The required derivative is $\frac{dp}{dq} = \frac{dp/dx}{dq/dx} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.

(B) Given $x=\log _{e}\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$.
Dividing numerator and denominator by $\cos \frac{y}{2}$,we get $e^x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}} = \tan \left(\frac{\pi}{4}-\frac{y}{2}\right) \quad \dots(i)$.
Differentiating both sides with respect to $x$,we get $e^x = \sec^2 \left(\frac{\pi}{4}-\frac{y}{2}\right) \cdot \left(-\frac{1}{2}\right) \frac{dy}{dx}$.
Thus,$\frac{dy}{dx} = -2e^x \cos^2 \left(\frac{\pi}{4}-\frac{y}{2}\right)$.
When $t=\frac{1}{2}$,$\tan \frac{y}{2} = \sqrt{\frac{1-1/2}{1+1/2}} = \sqrt{\frac{1/2}{3/2}} = \frac{1}{\sqrt{3}}$.
So,$\frac{y}{2} = \frac{\pi}{6}$.
Substituting $\frac{y}{2} = \frac{\pi}{6}$ into equation $(i)$,$e^x = \tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right) = \tan \frac{\pi}{12} = 2-\sqrt{3}$.
Now,$\cos^2 \left(\frac{\pi}{4}-\frac{\pi}{6}\right) = \cos^2 \frac{\pi}{12} = \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)^2 = \frac{3+1+2\sqrt{3}}{8} = \frac{4+2\sqrt{3}}{8} = \frac{2+\sqrt{3}}{4}$.
Therefore,$\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}} = -2(2-\sqrt{3}) \cdot \frac{2+\sqrt{3}}{4} = -\frac{1}{2}(4-3) = -\frac{1}{2}$.

(B) Let $x = a \cos^3 t$ and $y = a \sin^3 t$. We need to find $\frac{d^2 y}{dx^2}$.
First,find $\frac{dy}{dt} = 3a \sin^2 t \cos t$ and $\frac{dx}{dt} = -3a \cos^2 t \sin t$.
Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\tan t$.
Now,find the second derivative $\frac{d^2 y}{dx^2} = \frac{d}{dx}(-\tan t) = \frac{d}{dt}(-\tan t) \cdot \frac{dt}{dx}$.
Since $\frac{dx}{dt} = -3a \cos^2 t \sin t$,then $\frac{dt}{dx} = \frac{1}{-3a \cos^2 t \sin t}$.
So,$\frac{d^2 y}{dx^2} = (-\sec^2 t) \cdot \frac{1}{-3a \cos^2 t \sin t} = \frac{1}{3a \cos^4 t \sin t}$.
At $t = \frac{\pi}{4}$,$\cos t = \frac{1}{\sqrt{2}}$ and $\sin t = \frac{1}{\sqrt{2}}$.
$\frac{d^2 y}{dx^2} = \frac{1}{3a (\frac{1}{\sqrt{2}})^4 (\frac{1}{\sqrt{2}})} = \frac{1}{3a (\frac{1}{4}) (\frac{1}{\sqrt{2}})} = \frac{4 \sqrt{2}}{3a}$.

(A) Given $x = \log t$ and $y + 1 = \frac{1}{t}$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = -\frac{1}{t^{2}}$.
Now,find $\frac{dx}{dy}$ using the chain rule:
$\frac{dx}{dy} = \frac{dx/dt}{dy/dt} = \frac{1/t}{-1/t^{2}} = -t$.
Next,find the second derivative $\frac{d^{2}x}{dy^{2}}$:
$\frac{d^{2}x}{dy^{2}} = \frac{d}{dy}(-t) = \frac{d}{dt}(-t) \cdot \frac{dt}{dy} = (-1) \cdot \frac{1}{dy/dt} = (-1) \cdot \frac{1}{-1/t^{2}} = t^{2}$.
Also,calculate $e^{-x}$:
$e^{-x} = e^{-\log t} = e^{\log(t^{-1})} = \frac{1}{t}$.
Finally,substitute these values into the expression $e^{-x} \frac{d^{2}x}{dy^{2}} + \frac{dx}{dy}$:
$\left(\frac{1}{t}\right)(t^{2}) + (-t) = t - t = 0$.

(B) Let $u = \cot ^{-1} x$ and $v = \log (1+x^{2})$.
First,we find the derivative of $u$ with respect to $x$:
$\frac{du}{dx} = -\frac{1}{1+x^{2}}$.
Next,we find the derivative of $v$ with respect to $x$ using the chain rule:
$\frac{dv}{dx} = \frac{1}{1+x^{2}} \times \frac{d}{dx}(1+x^{2}) = \frac{2x}{1+x^{2}}$.
Now,we find the derivative of $u$ with respect to $v$:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-1/(1+x^{2})}{2x/(1+x^{2})} = -\frac{1}{2x}$.
Thus,the correct option is $B$.

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

(B) Let $y_1 = \log (\sec \theta + \tan \theta)$.
Then,$\frac{dy_1}{d\theta} = \frac{1}{\sec \theta + \tan \theta} \cdot (\sec \theta \tan \theta + \sec^2 \theta)$.
$\frac{dy_1}{d\theta} = \frac{\sec \theta (\tan \theta + \sec \theta)}{\sec \theta + \tan \theta} = \sec \theta$.
Let $y_2 = \sec \theta$.
Then,$\frac{dy_2}{d\theta} = \sec \theta \tan \theta$.
We need to find $\frac{dy_1}{dy_2} = \frac{dy_1/d\theta}{dy_2/d\theta} = \frac{\sec \theta}{\sec \theta \tan \theta} = \frac{1}{\tan \theta} = \cot \theta$.
At $\theta = \frac{\pi}{4}$,$\frac{dy_1}{dy_2} = \cot \frac{\pi}{4} = 1$.

(D) Given,$x=f(t)$ and $y=g(t)$.
First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dx}{dt} = f'(t)$ and $\frac{dy}{dt} = g'(t)$.
Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$.
Now,we find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g'(t)}{f'(t)} \right) = \frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) \cdot \frac{dt}{dx}$.
Using the quotient rule for differentiation:
$\frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) = \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^2}$.
Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{f'(t)}$,we have:
$\frac{d^2y}{dx^2} = \left[ \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^2} \right] \cdot \frac{1}{f'(t)} = \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^3}$.

(A) Let $u = \cos^{3} x$ and $v = \sin^{3} x$.
Applying the chain rule,we find the derivatives with respect to $x$:
$\frac{du}{dx} = 3 \cos^{2} x (-\sin x) = -3 \cos^{2} x \sin x$
$\frac{dv}{dx} = 3 \sin^{2} x (\cos x) = 3 \sin^{2} x \cos x$
Now,the derivative of $u$ with respect to $v$ is given by:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-3 \cos^{2} x \sin x}{3 \sin^{2} x \cos x}$
Simplifying the expression:
$\frac{du}{dv} = -\frac{\cos x}{\sin x} = -\cot x$

(A) Let $y = \sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$.
Put $t = \tan \theta$,then $\theta = \tan ^{-1} t$.
$y = \sin ^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right) = \sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) = \sin ^{-1}(\sin \theta) = \theta = \tan ^{-1} t$.
Let $z = \cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$.
$z = \cos ^{-1}\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right) = \cos ^{-1}\left(\frac{1}{\sec \theta}\right) = \cos ^{-1}(\cos \theta) = \theta = \tan ^{-1} t$.
Since $y = \theta$ and $z = \theta$,we have $y = z$.
Therefore,$\frac{dy}{dz} = \frac{d}{dz}(z) = 1$.

(B) Given $x = a(1 - \cos \theta)$ and $y = a(\theta - \sin \theta)$.
First,find $\frac{dx}{d\theta} = a \sin \theta$ and $\frac{dy}{d\theta} = a(1 - \cos \theta)$.
Then,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1 - \cos \theta)}{a \sin \theta} = \frac{2 \sin^{2}(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \tan(\theta/2)$.
Now,differentiate with respect to $x$: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\tan(\theta/2)) = \sec^{2}(\theta/2) \cdot \frac{1}{2} \cdot \frac{d\theta}{dx}$.
Since $\frac{dx}{d\theta} = a \sin \theta$,we have $\frac{d\theta}{dx} = \frac{1}{a \sin \theta}$.
Substituting this: $\frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sec^{2}(\theta/2) \cdot \frac{1}{a(2 \sin(\theta/2) \cos(\theta/2))} = \frac{1}{4a \sin(\theta/2) \cos^{3}(\theta/2)} = \frac{1}{4a \sin(\theta/2) \cos(\theta/2) \cos^{2}(\theta/2)} = \frac{1}{2a \sin \theta \cos^{2}(\theta/2)}$.
Alternatively,$\frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sec^{2}(\theta/2) \cdot \frac{1}{a(2 \sin(\theta/2) \cos(\theta/2))} = \frac{1}{4a \sin(\theta/2) \cos^{3}(\theta/2)} = \frac{\sec^{4}(\theta/2)}{4a}$.

(B) Given $x = \sin t$,we have $\frac{dx}{dt} = \cos t$.
Given $y = a e^{t \sqrt{2}} + b e^{-t \sqrt{2}}$.
Then $\frac{dy}{dt} = a \sqrt{2} e^{t \sqrt{2}} - b \sqrt{2} e^{-t \sqrt{2}} = \sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})}{\cos t}$.
Let $y' = \frac{dy}{dx}$. Then $y' \cos t = \sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})$.
Differentiating both sides with respect to $t$:
$y'' \cos t - y' \sin t = \sqrt{2} \cdot \sqrt{2}(a e^{t \sqrt{2}} + b e^{-t \sqrt{2}}) = 2y$.
Since $\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$,we have $y'' \sqrt{1 - x^2} - y' \frac{dx}{dt} \cdot \frac{1}{\cos t} \cdot \sin t = 2y$.
Actually,using $y' = \frac{dy}{dx}$,we have $\frac{d}{dx}(y' \cos t) = \frac{d}{dx}(\sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}}))$.
$y'' \cos t + y'(-\sin t) \frac{dt}{dx} = 2y \frac{dt}{dx}$.
Since $\frac{dt}{dx} = \frac{1}{\cos t}$,we get $y'' \cos t - y' \frac{\sin t}{\cos t} = 2y \frac{1}{\cos t}$.
Multiplying by $\cos t$: $y'' \cos^2 t - y' \sin t = 2y$.
Substituting $\cos^2 t = 1 - x^2$ and $\sin t = x$:
$(1 - x^2) y'' - x y' = 2y$.
Comparing with $(1 - x^2) y'' - x y' = ky$,we get $k = 2$.

(B) Given $x = a(1 - \cos \theta)$ and $y = a(\theta + \sin \theta)$.
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a(0 - (-\sin \theta)) = a \sin \theta$.
Next,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a(1 + \cos \theta)$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1 + \cos \theta)}{a \sin \theta} = \frac{1 + \cos \theta}{\sin \theta}$.
Using trigonometric identities $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$\frac{dy}{dx} = \frac{2 \cos^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = \cot \frac{\theta}{2}$.
Thus,the correct option is $B$.

(A) Given $x = a \cos \theta$ and $y = a \sin \theta$.
First,find $\frac{dx}{d\theta} = -a \sin \theta$ and $\frac{dy}{d\theta} = a \cos \theta$.
Then,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(-\cot \theta) = \frac{d}{d\theta}(-\cot \theta) \cdot \frac{d\theta}{dx}$.
Since $\frac{d}{d\theta}(-\cot \theta) = \operatorname{cosec}^2 \theta$ and $\frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{-a \sin \theta}$.
Therefore,$\frac{d^2 y}{dx^2} = \operatorname{cosec}^2 \theta \cdot \left( -\frac{1}{a \sin \theta} \right) = -\frac{1}{a} \operatorname{cosec}^3 \theta$.

(C) Given $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$.
Multiply $x$ and $y$:
$xy = \sqrt{10^{\sin^{-1} t} \cdot 10^{\cos^{-1} t}} = \sqrt{10^{\sin^{-1} t + \cos^{-1} t}}$.
Since $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have:
$xy = \sqrt{10^{\pi/2}}$.
Differentiating both sides with respect to $x$:
$x \frac{dy}{dx} + y(1) = 0$.
$x \frac{dy}{dx} = -y$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(D) Given $x = at^2$ and $y = 2at$.
First,find $\frac{dx}{dy}$ using the chain rule:
$\frac{dx}{dy} = \frac{dx/dt}{dy/dt}$.
Since $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$,
$\frac{dx}{dy} = \frac{2at}{2a} = t$.
Now,differentiate $\frac{dx}{dy}$ with respect to $y$:
$\frac{d^2 x}{dy^2} = \frac{d}{dy}(t) = \frac{d}{dt}(t) \times \frac{dt}{dy}$.
Since $\frac{dt}{dy} = \frac{1}{dy/dt} = \frac{1}{2a}$,
$\frac{d^2 x}{dy^2} = 1 \times \frac{1}{2a} = \frac{1}{2a}$.
Thus,the correct option is $D$.

(C) Let $u = \sin ^2 x$ and $v = \cos ^2 x$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
First,differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(\sin ^2 x) = 2 \sin x \cos x = \sin(2x)$.
Next,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx}(\cos ^2 x) = 2 \cos x (-\sin x) = -2 \sin x \cos x = -\sin(2x)$.
Now,calculate the derivative:
$\frac{du}{dv} = \frac{\sin(2x)}{-\sin(2x)} = -1$.

(B) Given $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$.
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a(1 + \cos \theta)$.
Next,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a(0 - (-\sin \theta)) = a \sin \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}$.
Using trigonometric identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$:
$\frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = \tan \frac{\theta}{2}$.
Thus,the correct option is $B$.

(C) Given $x = a(1 - \cos \theta)$ and $y = a(\theta + \sin \theta)$.
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a(0 - (-\sin \theta)) = a \sin \theta$.
Next,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a(1 + \cos \theta)$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1 + \cos \theta)}{a \sin \theta} = \frac{1 + \cos \theta}{\sin \theta}$.
Using trigonometric identities $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$\frac{dy}{dx} = \frac{2 \cos^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = \cot \frac{\theta}{2}$.
Thus,the correct option is $C$.

(D) Given parametric equations are $x = at^2$ and $y = 2at$.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$.
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$.
Using the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at}$.
Since $t \neq 0$ and $a \neq 0$,we simplify the expression:
$\frac{dy}{dx} = \frac{1}{t}$.
Therefore,the correct option is $D$.

(A) Given $x = at^2$ and $y = 2at$.
First,find the first derivative $\frac{dy}{dx}$:
$\frac{dx}{dt} = 2at$
$\frac{dy}{dt} = 2a$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} = t^{-1}$.
Now,find the second derivative $y_2 = \frac{d^2y}{dx^2}$:
$y_2 = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}$
$y_2 = \frac{d}{dt}(t^{-1}) \cdot \frac{1}{2at}$
$y_2 = (-t^{-2}) \cdot \frac{1}{2at} = -\frac{1}{t^2} \cdot \frac{1}{2at} = -\frac{1}{2at^3}$.
Thus,the correct option is $A$.

(C) Given that $x = at^2$ and $y = 2at$.
We need to find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$.
Differentiating $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$.
Now,$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.
Since $y = 2at$,we have $t = \frac{y}{2a}$.
Substituting $t$ in the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{y/(2a)} = \frac{2a}{y}$.
Alternatively,from $x = at^2$,we have $t^2 = \frac{x}{a}$,so $t = \sqrt{\frac{x}{a}}$.
Thus,$\frac{dy}{dx} = \frac{1}{\sqrt{x/a}} = \sqrt{\frac{a}{x}}$.
Looking at the options,if we express $\frac{1}{t}$ in terms of $x$ and $y$,we note that $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
Given $y = 2at$,then $t = \frac{y}{2a}$.
Substituting $t$ into $\frac{dy}{dx} = \frac{1}{t}$ gives $\frac{2a}{y}$.
However,checking the ratio $\frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
If we evaluate $\frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
Comparing with options,$\frac{1}{t}$ is the derivative. If the question implies expressing in terms of $x$ and $y$,$\frac{dy}{dx} = \frac{2a}{y}$.

(C) Given $x = at^2$ and $y = 2at$.
First,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$.
Next,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$.
Now,use the chain rule for parametric differentiation:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.
Since $t = \frac{y}{2a}$,we can also express this as $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
Alternatively,from $y = 2at$,we have $t = \frac{y}{2a}$.
Substituting $t$ into $x = at^2$:
$x = a(\frac{y}{2a})^2 = a(\frac{y^2}{4a^2}) = \frac{y^2}{4a}$.
$y^2 = 4ax$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 4a$.
$\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$.
Substituting $y = 2at$:
$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.

(A) Given that,$\sin x = \frac{2t}{1+t^2}$ and $\tan y = \frac{2t}{1-t^2}$.
We know the standard trigonometric substitutions:
$x = \sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1}t$ (for $|t| \le 1$).
$y = \tan^{-1}\left(\frac{2t}{1-t^2}\right) = 2\tan^{-1}t$ (for $|t| < 1$).
Since $x = 2\tan^{-1}t$ and $y = 2\tan^{-1}t$,it follows that $x = y$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

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