Q.If $\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)$, prove that $\dfrac{dy}{dx} = \sqrt{\dfrac{1 - y^2}{1 - x^2}}$.

Concept: Implicit Differentiation — differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate term by term.

For $\sqrt{1 - x^2}$, derivative is $\frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}}$.

For $\sqrt{1 - y^2}$, derivative is $-\frac{y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx}$.

Right-hand side derivative: $a(1 - \frac{dy}{dx})$.

Step 2: So we have

Step 3: Collect terms with $\frac{dy}{dx}$:

$\frac{dy}{dx} \left( a - \frac{y}{\sqrt{1 - y^2}} \right) = a + \frac{x}{\sqrt{1 - x^2}}.$ …

The problem asks us to prove a relationship between the derivatives of $x$ and $y$ that are linked by an equation involving square roots. The direct approach — differentiating term by term — will work, but we need to handle the square roots carefully. The key insight is that expressions like $\sqrt{1 - x^2}$ naturally suggest a trigonometric substitution: let $x = \sin \theta$ and $y = \sin \phi$. This turns the messy radicals into simple cosines, and the equation becomes a clean trigonometric identity. Then differentiating becomes straightforward.

Let’s work through it step by step.

1. Set up the trigonometric substitution Since $1 - x^2$ and $1 - y^2$ appear under square roots, we assume $|x| \leq 1$ and $|y| \leq 1$ (so the radicals are real). Let

where $\theta, \phi \in [-\pi/2, \pi/2]$ to keep the principal values. Then

(positive because cosine is non-negative on that interval).

2. Rewrite the given equation The equation becomes:

3. Use sum-to-product identities Recall:

Substituting:

4. Simplify the equation If $\cos\frac{\theta+\phi}{2} \neq 0$, we can cancel $2 \cos\frac{\theta+\phi}{2}$ from both sides:

This gives:

So $\frac{\theta-\phi}{2}$ is constant (since $a$ is constant). Hence $\theta - \phi$ is constant.

5. Differentiate the relation $\theta - \phi = \text{constant}$ …

Method: Implicit Differentiation of an Equation Involving Square-Root Terms

Use this method whenever $y$ is tangled up with $x$ inside radicals or transcendental expressions and you are asked to find or prove a formula for $\dfrac{dy}{dx}$.

Steps

Step 1: Differentiate both sides with respect to $x$, treating $y$ as $y(x)$

Every term containing only $x$ differentiates normally. Every term containing $y$ picks up an extra factor of $\dfrac{dy}{dx}$ by the chain rule — e.g. for a radical, $\dfrac{d}{dx}\sqrt{1-y^2}=\dfrac{-y}{\sqrt{1-y^2}}\cdot\dfrac{dy}{dx}$.

Step 2: Collect all $\dfrac{dy}{dx}$ terms on one side

Move every term containing $\dfrac{dy}{dx}$ to one side, everything else to the other, then factor it out:

Step 3: Solve for $\dfrac{dy}{dx}$

Divide by the factored coefficient. At this stage the result is usually a fraction in both $x$ and $y$ — this is expected, since $y$ was never isolated to begin with.

Step 4: Simplify using the original relation …

Common Mistakes

Mistake 1: Differentiating $\sqrt{1-y^2}$ as if $y$ were an ordinary variable

Why it's wrong: $y$ is a function of $x$, so differentiating $\sqrt{1-y^2}$ needs the chain rule — it must carry a factor of $\dfrac{dy}{dx}$: $\dfrac{d}{dx}\sqrt{1-y^2} = -\dfrac{y}{\sqrt{1-y^2}}\cdot\dfrac{dy}{dx}$. Dropping that factor leaves an equation with no $\dfrac{dy}{dx}$ to solve for. Correct approach: treat every $y$-term as $y(x)$ and attach $\dfrac{dy}{dx}$ the instant you differentiate it.

Mistake 2: Forgetting to differentiate the right side as a product

Why it's wrong: the right side is $a(x-y)$; its derivative is $a\left(1-\dfrac{dy}{dx}\right)$, not simply $a$. Skipping the $-a\dfrac{dy}{dx}$ piece removes one of the two places $\dfrac{dy}{dx}$ should appear, so the later "collect $\dfrac{dy}{dx}$ terms" step comes out wrong. Correct approach: differentiate the whole right side term by term before rearranging.

Mistake 3: Trying to finish without eliminating $a$ …