As per the given question,
If $\sin x=\frac{2 t}{1+t^{2}}, \tan y=\frac{2 t}{1-t^{2}}$ find $\frac{d y}{d x} \quad$
As per the given question,
If $x=3 \sin t-\sin 3 t, y=3 \cos t-\cos 3 t$, find $\frac{d y}{d x}$ at $t=\frac{\pi}{3}$.
As per the given question,
If $x=\frac{1+\log t}{t^{2}}, y=\frac{3+2 \log t}{t}$, find $\frac{d y}{d x}$
As per the given question,
If $x=\cos t\left(3-2 \cos ^{2}\right)$ and $y=\sin t\left(3-2 \sin ^{2} t\right)$ find the value of $\frac{d y}{d x}$ at $t=\frac{\pi}{4}$
As per the given question,
If $x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$, show that at $t=\frac{\pi}{4}, \frac{d y}{d x}=\frac{b}{a}$.
As per the given question,
If $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$ find $\frac{d y}{d x}$, at $\theta=\frac{\pi}{3}$.
As per the given question,
Find $dy/dx$, if $y\;=\;12(1-cos t),$ $x\;=\;10(1-sin t)$
As per the given question,
If $x=a\left(\frac{1+t^{2}}{1-t^{2}}\right)$ and $y=\frac{2 t}{1-t^{2}}$, find $\frac{d y}{d x} \quad$
As per the given question,
If $x=\left(t+\frac{1}{t}\right)^{2}$ and $y=a^{1+\frac{1}{t}}$, find $\frac{d y}{d x}$
As per the given question,
If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$. $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$
As per the given question,
If $x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)$ and $y=\tan ^{-1}\left(\frac{2 t}{1+t^{2}}\right) .-1
As per the given question,
If $x=a\left(t+\frac{1}{t}\right)$ and $y=a\left(t-\frac{1}{t}\right)$, prove that $\frac{d y}{d x}=\frac{x}{y}$.
As per the given question,
If $x=\cos t$ and $y=\sin t$, prove that $\frac{d y}{d x}=\frac{1}{\sqrt{3}}$ at $t=\frac{2 \pi}{3}$
As per the given question,
If $x=x=e^{\cos 2 t}$ and $y=e^{\sin 2 t}$, prove that $\frac{d y}{d x}=-\frac{y \log x}{x \log y}$
As per the given question,
If $x=2 \cos \theta-\infty \sin \theta$ and $y=2 \sin \theta-\sin 2 \theta$, prove that $\frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right)$
As per the given question,
Find $\frac{d y}{d x}$, when ${x}=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$
As per the given question,
Find $\frac{d y}{d x}$, when $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$ and $y=\sin ^{-1} \frac{t}{\sqrt{1+t^{2}}}, t \in R$
As per the given question,
Find $\frac{d y}{d x}$, when $x=\frac{2 t}{1+t^{2}}$ and $y=\frac{1-t^{2}}{1+t^{2}}$
As per the given question,
Find $\frac{d y}{d x}$, when Find $\frac{d y}{d x}$, when $x=e^{\theta}\left(\theta+\frac{1}{\theta}\right)$ and $y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$
As per the given question,
If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$. $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$
As per the given question,
Find $\frac{d y}{d x}$, when $x=\frac{3 a t}{1+t^{2}}$ and $y=\frac{3 a t^{2}}{1+t^{2}}$
As per the given question,
Find $\frac{d y}{d x}$, when $\quad x=\frac{e^{t}+e^{-t}}{2}$ and $y=\frac{e^{t}-e^{-t}}{2}$
As per the given question,
Find $\frac{d y}{d x}$, when $x=a(1-\cos \theta)$ and $y=a(\theta+\sin \theta) \mathrm{at} \theta=\frac{\pi}{2} .$
As per the given question,
Find $\frac{d y}{d x}$, when $\quad x=b \sin ^{2} \theta$ and $y=a \cos ^{2} \theta$
As per the given question,
Find $\frac{d y}{d x}$, when $x=a e^{\theta}(\sin \theta-\cos \theta), y=a e^{\theta}(\sin \theta+\cos \theta)$
As per the given question,
Find $\frac{d y}{d x}$, when $\quad x=a \cos \theta, y=b \sin \theta$
As per the given question,
Find $\frac{d y}{d x}$, when $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$
As per the given question,
Find $dy/dx$ when: $x\;=\;at^2$ and $y\;=\;2at$
We have, $y=2 a t$ $\frac{d y}{d t}=2 a \frac{d}{d t}(t)=2 a(1)=2 a \\ \text { also } x=a t^{2} \\ \frac{d x}{d x}=a \frac{d}{d t}\left(t^{2}\right)=a(2 t)=2 a t \\ \text { now } \frac{d y}{d...
If $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta$, prove that $\frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right)$.
Find dy/dx, when $x=\frac{1-t^{2}}{1+t^{2}} \text { and } y=\frac{2 t}{1+t^{2}}$
Find dy/dx, when $x=\cos ^{-1} \frac{1}{\sqrt{1+t^{2}}} \text { and } y=\sin ^{-1} \frac{1}{\sqrt{1+t^{2}}}, t \epsilon R$
Find dy/dx, when x = 2 t / 1+t^2 and y = 1-t^2 / 1+t^2.
Given, $x=2 t /\left(1+t^{2}\right)$ On differentiating $x$ with respect to t using quotient rule, $$ \begin{array}{l} \frac{\mathrm{dx}}{\mathrm{dt}}=\left[\frac{\left(1+\mathrm{t}^{2}\right)...