If x = \sqrt 2^\cos ec^ - 1 and y = \sqrt 2^sec^ - 1t \,\,( | t | \ge 1 ), then dy \over dx is equal to :

Differentiation PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q28 with Solution

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Correct answer: C

The notation in the question is clearly intended as:

$x=\sqrt{2^{\cosec^{-1} t}}, \qquad y=\sqrt{2^{\sec^{-1} t}} \quad (|t|\ge 1)$

So we can rewrite them as:

$x=2^{\frac{1}{2}\cosec^{-1} t}, \qquad y=2^{\frac{1}{2}\sec^{-1} t}$

We need to find $\dfrac{dy}{dx}$ .

Using

$\frac{d}{dt}(2^{u})=2^{u}\ln 2\cdot \frac{du}{dt},$

we get

$\frac{dx}{dt}=x\ln 2\cdot \frac{1}{2}\frac{d}{dt}(\cosec^{-1} t).$

Now,

$\frac{d}{dt}(\cosec^{-1} t)=-\frac{1}{|t|\sqrt{t^2-1}}.$

Hence,

=-\frac{x\ln 2}{2|t|\sqrt{t^2-1}}.$$ --- 3. **Differentiate $y$ with respect to $t$** Similarly, $$\frac{dy}{dt}=y\ln 2\cdot \frac{1}{2}\frac{d}{dt}(\sec^{-1} t).$$ Using $$\frac{d}{dt}(\sec^{-1} t)=\frac{1}{|t|\sqrt{t^2-1}},$$ we get $$\frac{dy}{dt}=\frac{y\ln 2}{2|t|\sqrt{t^2-1}}.$$ --- 4. **Compute $\dfrac{dy}{dx}$** $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{\frac{y\ln 2}{2|t|\sqrt{t^2-1}}}{-\frac{x\ln 2}{2|t|\sqrt{t^2-1}}} =-\frac{y}{x}.$$ --- 5. **Match with the options** $$\frac{dy}{dx}=-\frac{y}{x}$$ So the correct option is: **C: $-\dfrac{y}{x}$**