Animated Solution for Mathematics - Differentiation: Let $y=y(x)$ be a function of $x$ satisfying $y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$ where $k$ is a constant and $y\left(\frac{1}{2}\right)=-\frac{1}{4}$. Then $\frac{dy}{dx}$ at $x=\frac{1}{2}$, is equal to :

Imagine you are standing before a complex, intimidating equation: $y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$. At first glance, it looks like a tangled mess of variables, square roots, and constants.

A student might be tempted to dive headfirst into the product rule and chain rule, creating a sprawling algebraic landscape that is easy to get lost in. But wait—take a breath.

In mathematics, as in life, sometimes the most complex problems are just simple truths wearing a disguise. Our goal is to find $\frac{dy}{dx}$ at $x=\frac{1}{2}$, and we are going to do it with elegance.

First, let's bring order to the chaos. By rearranging the terms, we get:

Now, look at the structure. Does it feel familiar? It should. It is a classic pattern that screams for trigonometric substitution.

Let $x=\sin A$ and $y=\sin B$. This implies $A=\arcsin x$ and $B=\arcsin y$.

Why do we do this? Because the square root terms $\sqrt{1-{\sin }^2\theta }$ transform beautifully into $\cos \theta$. Our equation becomes:

Suddenly, the fog lifts. This is the exact expansion of the compound angle formula: $\sin (A+B)=\sin A\cos B+\cos A\sin B$. The entire equation collapses into the stunningly simple:

Now, we take the inverse sine of both sides: $A+B=\arcsin k$. Since $k$ is a constant, $\arcsin k$ is also a constant—let's call it $C$.

Substituting back our original variables, we arrive at:

This is the 'soul' of the problem. We have stripped away the layers of complexity to reveal a simple relationship between $x$ and $y$.

Now, differentiating with respect to $x$ becomes a trivial task. The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$, and the derivative of $\arcsin y$ is $\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}$.

The derivative of the constant $C$ is, of course, zero. Thus:

Isolating $\frac{dy}{dx}$, we find:

We are given $x=\frac{1}{2}$ and $y=-\frac{1}{4}$. Let's calculate the components.

For the denominator:

For the numerator:

Plugging these into our derivative expression, we get:

Simplifying this, we multiply by the reciprocal:

The $\sqrt{3}$ terms cancel, the $2$ and $4$ simplify, and we are left with the elegant result: $-\frac{\sqrt{5}}{2}$. We have conquered the monster, not by brute force, but by recognizing the hidden harmony within the math.