Animated Solution for Mathematics - Differentiation: If $x=\sqrt{2^{{\text{cosec}}^{-1}t}}$ and $y=\sqrt{2^{{\text{sec}}^{-1}t}}$ ($|t|\ge 1$), then $\frac{dy}{dx}$ is equal to :

Welcome, future engineer. Today, we are going to tackle a problem that looks like a daunting calculus exercise but is actually a beautiful lesson in pattern recognition.

When you first look at the equations $x=\sqrt{2^{{\text{cosec}}^{-1}t}}$ and $y=\sqrt{2^{{\text{sec}}^{-1}t}}$, your instinct might be to reach for the chain rule. You might think, "I will find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ and then divide them."

While that is a valid path, it is a path filled with potential pitfalls. In the world of JEE Advanced, we don't just want to solve; we want to solve with elegance.

Look at the exponents. We have ${\text{cosec}}^{-1}t$ and ${\text{sec}}^{-1}t$. Does that ring a bell?

It should! One of the most powerful tools in your trigonometry toolkit is the identity ${\text{sec}}^{-1}t+{\text{cosec}}^{-1}t=\frac{\pi }{2}$ for $|t|\ge 1$. This identity is the key that unlocks the entire problem.

To use it, we need to bring those exponents together. How do we add exponents? By multiplying the bases! Let us multiply $x$ and $y$:

By combining them under a single radical, we get:

Using the exponent rule $a^m\cdot a^n=a^{m+n}$, the expression simplifies beautifully:

Now, substitute our identity into the exponent. The variable $t$ vanishes entirely!

Let $C=\sqrt{2^{\pi /2}}$. We have discovered that $x\cdot y=C$.

This is not just any equation; it is the equation of a rectangular hyperbola. Geometrically, this means that as you move along the curve, the product of your coordinates remains constant. It is a simple, elegant relationship that hides behind the complex-looking parametric form.

Now that we have $x\cdot y=C$, finding the derivative is a breeze. We use implicit differentiation with respect to $x$:

Applying the product rule on the left side, we get:

Since $\frac{dx}{dx}=1$, this simplifies to:

Solving for $\frac{dy}{dx}$, we arrive at our final answer:

This result is not just a collection of symbols; it is the slope of the tangent to the hyperbola at any point $(x,y)$. It is negative, confirming that the curve is decreasing in the first quadrant.

You see? By taking a moment to observe the structure of the problem, we turned a potential nightmare into a moment of clarity. Keep looking for these patterns—they are the secret language of physics and mathematics.