Q.If $x = \dfrac{1-t^2}{1+t^2}$, $y = \dfrac{2t}{1+t^2}$, then show that $\dfrac{dy}{dx} + \dfrac{x}{y} = 0$.

$\frac{dy}{dx} = \cot\frac{\theta}{2}$.

Differentiate the parametric equations with respect to $\theta$:

$\frac{dx}{d\theta} = a\sin\theta, \qquad \frac{dy}{d\theta} = a(1+\cos\theta).$

Then

$\frac{dy}{dx} = \frac{a(1+\cos\theta)}{a\sin\theta} = \frac{1+\cos\theta}{\sin\theta}.$

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For a curve given parametrically as $x=x(\theta)$, $y=y(\theta)$, use $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$.

Given: $x = a\cos\theta$, $y = a\sin\theta$

Differentiate $x$ w.r.t. $\theta$:

$\frac{dx}{d\theta} = -a\sin\theta$

Differentiate $y$ w.r.t. $\theta$:

$\frac{dy}{d\theta} = a\cos\theta$

Combine using the chain rule: …

Differentiate parametrically: $\tfrac{dy}{dx}=-t$, then $\tfrac{d^2y}{dx^2}=-\sqrt{1-t^2}$, giving $0$ at $t=1$.

Second-order parametric differentiation is a CBSE/NCERT Class 12 continuity and differentiability topic.

With $x=\sin^{-1}t$ and $y=\sqrt{1-t^2}$:

$\frac{dx}{dt} = \frac{1}{\sqrt{1-t^2}}, \qquad \frac{dy}{dt} = \frac{-t}{\sqrt{1-t^2}}.$

So

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-t/\sqrt{1-t^2}}{1/\sqrt{1-t^2}} = -t.$

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Take logarithms of both parametric equations, differentiate w.r.t. $t$, and divide $dy/dt$ by $dx/dt$.

Given $x = a^{\sin^{-1}t}$ and $y = a^{\cos^{-1}t}$.

Differentiate $x$ w.r.t. $t$: Take $\ln$ of both sides: $\ln x = (\sin^{-1}t)\ln a$.

Differentiating w.r.t. $t$:

$\dfrac{1}{x}\dfrac{dx}{dt} = \dfrac{\ln a}{\sqrt{1-t^2}} \implies \dfrac{dx}{dt} = \dfrac{x\ln a}{\sqrt{1-t^2}}$

Differentiate $y$ w.r.t. $t$: Take $\ln$: $\ln y = (\cos^{-1}t)\ln a$. …

For parametric equations, divide $dy/dt$ by $dx/dt$ (chain rule).

Given $x=f(t)$ and $y=g(t)$, both functions of the parameter $t$.

By the chain rule, provided $\dfrac{dx}{dt}\neq0$:

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{g'(t)}{f'(t)}$

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$\frac{dy}{dx} = \frac{b}{a}\operatorname{cosec}\theta$.

This is a parametric differentiation. Differentiate each with respect to $\theta$:

$\frac{dx}{d\theta} = a\sec\theta\tan\theta,\qquad \frac{dy}{d\theta} = b\sec^2\theta.$

Then …

For parametric equations, dy/dx is found as (dy/dt) divided by (dx/dt).

Given $x=4t$ so $\dfrac{dx}{dt}=4$, and $y=\dfrac{4}{t}=4t^{-1}$ so $\dfrac{dy}{dt}=-4t^{-2}=-\dfrac{4}{t^2}$.

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Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.

Given $x=at^2$, $y=2at$.

$\dfrac{dx}{dt}=2at, \qquad \dfrac{dy}{dt}=2a$ …

With $x=a\cos^{2}\theta,\ y=b\sin^{2}\theta$, $\dfrac{dy}{dx}=-\dfrac{b}{a}$.

Differentiate each with respect to $\theta$:

$\dfrac{dx}{d\theta}=a\cdot 2\cos\theta(-\sin\theta)=-a\sin 2\theta,$

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Use parametric differentiation: $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$.

$x=a\cos\theta \Rightarrow \dfrac{dx}{d\theta} = -a\sin\theta$. $y=a\sin\theta \Rightarrow \dfrac{dy}{d\theta} = a\cos\theta$.

$\dfrac{dy}{dx} = \dfrac{a\cos\theta}{-a\sin\theta} = -\dfrac{\cos\theta}{\sin\theta} = -\cot\theta$.

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For parametric curves, $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

Given $x = at^2$, $y = 2at$.

$\dfrac{dx}{dt} = 2at$, $\dfrac{dy}{dt} = 2a$

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With $x=a\cos\theta,\ y=b\cos\theta$, both derivatives w.r.t. $\theta$ share the factor $-\sin\theta$, giving $dy/dx = b/a$.

$\dfrac{dx}{d\theta} = -a\sin\theta$, $\dfrac{dy}{d\theta} = -b\sin\theta$

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