Question

If $x=\sqrt{2^{cose{c}^{-1}t}}$ and $y=\sqrt{2^{se{c}^{-1}t}}$ $(|t|\ge 1)$, then $\frac{dy}{dx}$ is equal to.

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

Answer 1

The correct option is B $-\frac{y}{x}$

$x=\sqrt{2^{cose{c}^{-1}t}}$ and $y=\sqrt{2^{se{c}^{-1}t}}$

$\frac{dx}{dt}=\frac{1}{2\sqrt{2^{cose{c}^{-1}t}}}{2}^{cose{c}^{-1}t}\log 2\cdot \frac{-1}{t\sqrt{t^2-1}}$

$\frac{dy}{dt}=\frac{1}{2\sqrt{2^{se{c}^{-1}t}}}{2}^{se{c}^{-1}t}\log 2\cdot \frac{1}{t\sqrt{t^2-1}}$

Thus $\frac{dy}{dx}=$ $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-\sqrt{2^{cose{c}^{-1}t}}}{\sqrt{2^{se{c}^{-1}t}}}\frac{2^{se{c}^{-1}t}}{2^{cose{c}^{-1}t}}$

$\frac{dy}{dx}=$ $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\sqrt{\frac{2^{se{c}^{-1}t}}{2^{cose{c}^{-1}t}}}=\frac{-y}{x}$