Question

If $x=\theta -\frac{1}{\theta },y=\theta +\frac{1}{\theta }$ then $\frac{dy}{dx}$=

• A. $\frac{xy}{y^2+2}$
• B. y/x
• C. -x/y
• D. -y/x

Answer 1

The correct option is A $\frac{xy}{y^2+2}$

$x=\theta -\frac{1}{\theta }$

$\frac{dx}{d\theta }=1-\left(\frac{-1}{{\theta }^2}\right)$

$\therefore \frac{dx}{d\theta }=1+\frac{1}{{\theta }^2}$

$y=\theta +\frac{1}{\theta }$

$\frac{dy}{d\theta }=1+\left(\frac{-1}{{\theta }^2}\right)$

$\therefore \frac{dy}{d\theta }=1-\frac{1}{{\theta }^2}$

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{1-\frac{1}{{\theta }^2}}{1+\frac{1}{{\theta }^2}}$

We have $x=\theta -\frac{1}{\theta }$ and $y=\theta +\frac{1}{\theta }$

$\Rightarrow xy={\theta }^2-\frac{1}{{\theta }^2}$

and $y^2={(\theta +\frac{1}{\theta })}^2$

$\Rightarrow y^2={\theta }^2+\frac{1}{{\theta }^2}-2\theta \frac{1}{\theta }$

$\Rightarrow y^2={\theta }^2+\frac{1}{{\theta }^2}-2$

$\Rightarrow y^2+2={\theta }^2+\frac{1}{{\theta }^2}$

$\therefore \frac{dy}{dx}=\frac{1-\frac{1}{{\theta }^2}}{1+\frac{1}{{\theta }^2}}$

$=\frac{xy}{y^2+2}$