Question

If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$, then $\frac{dy}{dx}=$

• A. $\frac{y}{x}$
• B. $\frac{x}{y}$
• C. $-\frac{x}{y}$
• D. $-\frac{y}{x}$

Answer 1

The correct option is C

$-\frac{x}{y}$

Explanation for the correct option:

Given that,

$x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$

Step 1: Put $\mathit{t}\mathbf{=}\mathbf{tan}\mathbf{}\mathit{\theta }$and simplify

$\theta ={\tan }^{-1}t$

$$\begin{gathered} x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta } \\ \Rightarrow x=\cos 2\theta \end{gathered}$$

$$\begin{gathered} y=\frac{2\tan \theta }{1+{\tan }^2\theta } \\ \Rightarrow y=\sin 2\theta \end{gathered}$$

$$\begin{gathered} x^2={\cos }^{2}2\theta \\ {y}^2={\sin }^{2}2\theta \end{gathered}$$

Step 2: Add both $x^2$ and $y^2$:

$$\begin{gathered} x^2+y^2={\cos }^{2}2\theta +{\sin }^{2}2\theta \\ \Rightarrow x^2+y^2=1 \end{gathered}$$

Step 3: Differentiate with respect to $x$ both sides:

$$\begin{gathered} 2x+2y\frac{dy}{dx}=0 \\ \Rightarrow 2y\frac{dy}{dx}=-2x \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{2y} \\ \Rightarrow \frac{dy}{dx}=-\frac{x}{y} \end{gathered}$$

Hence, Option (C) is the correct answer.