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

> [!TLDR] > The key is to substitute $x = \sin \theta$ and $y = \sin \phi$, which simplifies the given equation into a trigonometric identity. Differentiating the resulting relation gives the required derivative directly. We are given: $$\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)$$ We need to prove that $$\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}$$ The presence of $\sqrt{1 - x^2}$ and $\sqrt{1 - y^2}$ strongly suggests a trigonometric substitution. Why? Because $\sqrt{1 - \sin^2 \theta} = |\cos \theta|$, and if we restrict the domain appropriately, we can drop the absolute value. This turns the messy square roots into simple trigonometric functions, and the equation becomes a relation between angles — much easier to differentiate. --- 1. **Substitute** Let $x = \sin \theta$ and $y = \sin \phi$, where $\theta, \phi \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ so that $\cos \theta, \cos \phi \geq 0$. Then $$\sqrt{1 - x^2} = \cos \theta, \quad \sqrt{1 - y^2} = \cos \phi$$ 2. **Rewrite the given equation** The equation becomes: $$\cos \theta + \cos \phi = a(\sin \theta - \sin \phi)$$ 3. **Use sum-to-product identities** Recall: $$\cos \theta + \cos \phi = 2 \cos\frac{\theta + \phi}{2} \cos\frac{\theta - \phi}{2}$$ $$\sin \theta - \sin \phi = 2 \cos\frac{\theta + \phi}{2} \sin\frac{\theta - \phi}{2}$$ Substituting: $$2 \cos\frac{\theta + \phi}{2} \cos\frac{\theta - \phi}{2} = a \cdot 2 \cos\frac{\theta + \phi}{2} \sin\frac{\theta - \phi}{2}$$ 4. **Cancel the common factor** Assuming $\cos\frac{\theta + \phi}{2} \neq 0$, divide both sides by $2\cos\frac{\theta + \phi}{2}$: $$\cos\frac{\theta - \phi}{2} = a \sin\frac{\theta - \phi}{2}$$ Hence: $$\cot\frac{\theta - \phi}{2} = a \quad \text{or} \quad \frac{\theta - \phi}{2} = \cot^{-1} a$$ This means $\theta - \phi$ is constant (since $a$ is constant). So: $$\theta - \phi = \text{constant}$$ > [!NOTE] > This derivation assumes $\cos\frac{\theta + \phi}{2} \neq 0$

If √(1 - x²) + √(1 - y²) = a(x - y), prove that (dy)/(d — CBSE Class 12 Mathematics 2024 PYQ

Question

\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)

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The key is to substitute $x = \sin \theta$ and $y = \sin \phi$ , which simplifies the given equation into a trigonometric identity. Differentiating the resulting relation gives the required derivative directly.

We are given:

\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)

We need to prove that

\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}

The presence of $\sqrt{1 - x^2}$ and $\sqrt{1 - y^2}$ strongly suggests a trigonometric substitution. Why? Because $\sqrt{1 - \sin^2 \theta} = |\cos \theta|$ , and if we restrict the domain appropriately, we can drop the absolute value. This turns the messy square roots into simple trigonometric functions, and the equation becomes a relation between angles — much easier to differentiate.

Substitute Let $x = \sin \theta$ and $y = \sin \phi$ , where $\theta, \phi \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ so that $\cos \theta, \cos \phi \geq 0$ . Then

\sqrt{1 - x^2} = \cos \theta, \quad \sqrt{1 - y^2} = \cos \phi

Rewrite the given equation The equation becomes:

\cos \theta + \cos \phi = a(\sin \theta - \sin \phi)

Use sum-to-product identities Recall:

\cos \theta + \cos \phi = 2 \cos\frac{\theta + \phi}{2} \cos\frac{\theta - \phi}{2}

\sin \theta - \sin \phi = 2 \cos\frac{\theta + \phi}{2} \sin\frac{\theta - \phi}{2}

Substituting:

2 \cos\frac{\theta + \phi}{2} \cos\frac{\theta - \phi}{2} = a \cdot 2 \cos\frac{\theta + \phi}{2} \sin\frac{\theta - \phi}{2}

Cancel the common factor Assuming $\cos\frac{\theta + \phi}{2} \neq 0$ , divide both sides by $2\cos\frac{\theta + \phi}{2}$ :

\cos\frac{\theta - \phi}{2} = a \sin\frac{\theta - \phi}{2}

Hence:

\cot\frac{\theta - \phi}{2} = a \quad \text{or} \quad \frac{\theta - \phi}{2} = \cot^{-1} a

This means $\theta - \phi$ is constant (since $a$ is constant). So:

\theta - \phi = \text{constant}

Note

This derivation assumes $\cos\frac{\theta + \phi}{2} \neq 0$ , which holds on the domain under consideration.

Differentiate the relation Since $\theta - \phi = \text{constant}$ , differentiating with respect to $x$ :

\frac{d\theta}{dx} - \frac{d\phi}{dx} = 0 \quad \Rightarrow \quad \frac{d\theta}{dx} = \frac{d\phi}{dx}

Relate derivatives back to $x$ and $y$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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CBSE 2026Set 65/2/11 markMCQ
$e^{-x} + e^{-y} = 2$
›Reveal solutionSolution
To find $\frac{dy}{dx}$ for an implicitly defined function, we differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule. The result is $\boxed{-e^{y-x}}$ .

When an equation relates $x$ and $y$ but does not explicitly express $y$ as a function of $x$ (like $y = f(x)$ ), we use a technique called implicit differentiation to find $\frac{dy}{dx}$ . The core idea is that even though $y$ isn't isolated, it is still a function of $x$ .

This means that when we differentiate a term involving $y$ with respect to $x$ , we must apply the chain rule. For example, if we differentiate $g(y)$ with respect to $x$ , we get $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ . This $\frac{dy}{dx}$ term is crucial and often the source of errors if overlooked.

Let's apply this to the given equation.

Differentiate both sides of the equation with respect to $x$ . The given equation is $e^{-x} + e^{-y} = 2$ . We apply the derivative operator $\frac{d}{dx}$ to every term:

$\frac{d}{dx}(e^{-x}) + \frac{d}{dx}(e^{-y}) = \frac{d}{dx}(2)$

Evaluate each derivative.

For the first term, $\frac{d}{dx}(e^{-x})$ :

Using the chain rule, if $u = -x$ , then $\frac{du}{dx} = -1$ .

So, $\frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x}$ .

For the second term, $\frac{d}{dx}(e^{-y})$ :

This is where implicit differentiation comes in. We treat $y$ as a function of $x$ .

Using the chain rule, if $v = -y$ , then $\frac{dv}{dx} = \frac{d}{dx}(-y) = -1 \cdot \frac{dy}{dx}$ .

So, $\frac{d}{dx}(e^{-y}) = e^{-y} \cdot \frac{d}{dx}(-y) = e^{-y} \cdot \left(-\frac{dy}{dx}\right) = -e^{-y} \frac{dy}{dx}$ .

Watch out

A common mistake is to forget the $\frac{dy}{dx}$ term when differentiating expressions involving $y$ with respect to $x$ . Remember, $y$ is a function of $x$ .

For the right-hand side, $\frac{d}{dx}(2)$ : …
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CBSE 2026Set A1 markMCQ
$y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}}$
(a) $\frac{1}{2y-1}$
(b) $\frac{\cos x}{2y-1}$
(c) $\frac{\sin x}{2y-1}$
(d) $\frac{2y-1}{\cos x}$
›Reveal solutionSolution

$\frac{dy}{dx} = \frac{\cos x}{2y-1}$ .

The infinite nested radical satisfies $y = \sqrt{\sin x + y}$ , so

$y^2 = \sin x + y.$

Differentiate both sides implicitly with respect to $x$ :

$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$

Collect $\frac{dy}{dx}$ : …

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CBSE 2026Set A1 markMCQ
$x^n + y^n = a^n$
(a) $-\frac{x^{n-1}}{y^{n-1}}$
(b) $\frac{x^{n-1}}{y^{n-1}}$
(c) $-\frac{y^{n-1}}{x^{n-1}}$
(d) $nx^{n-1}$
›Reveal solutionSolution

$\frac{dy}{dx} = -\frac{x^{n-1}}{y^{n-1}}$ .

Differentiate $x^n + y^n = a^n$ implicitly ( $a$ constant):

$n x^{n-1} + n y^{n-1}\frac{dy}{dx} = 0.$

Solve: …

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CBSE 2026Set ANNUAL1 markMCQ
$2x + 3y = \sin y$
(a) $\frac{3}{\sin y - 2}$
(b) $\frac{2}{\cos y - 3}$
(c) $\frac{\cos y + 3}{2}$
(d) $\frac{2}{\cos y}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ , then solve for $dy/dx$ .

$2x+3y=\sin y$

Differentiating: $2 + 3\dfrac{dy}{dx} = \cos y\,\dfrac{dy}{dx}$

$2 = \dfrac{dy}{dx}(\cos y - 3)$

…

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CBSE 2026Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides of $2x+3y=\sin y$ with respect to $x$ (using the chain rule for the $y$ -terms), then collect $\dfrac{dy}{dx}$ on one side.

Given: $2x + 3y = \sin y$

Differentiate both sides w.r.t. $x$ :

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$

$2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$

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

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CBSE 2025Set ANNUAL1 markMCQ
$x^2+y^2=2$
(a) $\dfrac{1-2x}{2y}$
(b) $\dfrac{2y}{1-2x}$
(c) $-\dfrac{x}{y}$
(d) $-\dfrac{y}{x}$
›Reveal solutionSolution

Differentiate $x^2+y^2=2$ implicitly with respect to $x$ , treating $y$ as a function of $x$ .

$\dfrac{d}{dx}(x^2+y^2)=\dfrac{d}{dx}(2)$

$2x+2y\dfrac{dy}{dx}=0$ …

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CBSE 2025Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .

Start from $ax+by^2=\cos y$ and differentiate w.r.t. $x$ :

$\frac{d}{dx}(ax)+\frac{d}{dx}(by^2)=\frac{d}{dx}(\cos y)$

$a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}.$

Gather the $\dfrac{dy}{dx}$ terms:

$2by\frac{dy}{dx}+\sin y\frac{dy}{dx}=-a$

$\frac{dy}{dx}\big(2by+\sin y\big)=-a.$

Hence

$\frac{dy}{dx}=\frac{-a}{2by+\sin y}.$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The value of dy/dx at (4, 1) of y³ - \sqrtx = 5 is ......................$
(a) 5/4
(b) 1/12
(c) 1/24
(d) 1/3
›Reveal solutionSolution

Differentiate the implicit relation $y^3 - \sqrt{x} = 5$ term by term with respect to $x$ , then substitute the point $(4, 1)$ .

Given: $y^3 - \sqrt{x} = 5$

Step 1 — differentiate implicitly w.r.t. $x$ :

$3y^2 \frac{dy}{dx} - \frac{1}{2\sqrt{x}} = 0$

Step 2 — solve for $dy/dx$ :

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}\cdot 3y^2} = \frac{1}{6y^2\sqrt{x}}$

…

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CBSE 2024Set D1 markMCQ
$y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \text{ to } \infty}}}$
(a) $\frac{\sin x}{2y - 1}$
(b) $\frac{\cos x}{y - 1}$
(c) $\frac{\cos x}{2y - 1}$
(d) $\frac{1}{2y - 1}$
›Reveal solutionSolution

$\frac{dy}{dx} = \frac{\cos x}{2y - 1}$ .

The infinite nested radical satisfies $y = \sqrt{\sin x + y}$ because the expression inside the outer root repeats. Square both sides:

$y^2 = \sin x + y.$

Differentiate implicitly with respect to $x$ :

$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$ …

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CBSE 2024Set ANNUAL1 markMCQ
$2x + 8y = \sin x$
(a) $\dfrac{\sin x - 2}{8}$
(b) $\dfrac{\cos x - 2}{8}$
(c) $\dfrac{\cos x + 2}{2}$
(d) $\dfrac{\cos x + 2}{3}$
›Reveal solutionSolution

Differentiate both sides of $2x+8y=\sin x$ implicitly with respect to $x$ and isolate $\dfrac{dy}{dx}$ .

$2x + 8y = \sin x$

Differentiating both sides w.r.t. $x$ :

$2 + 8\dfrac{dy}{dx} = \cos x$

…

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CBSE 2024Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ (implicit differentiation) and solve for $\dfrac{dy}{dx}$ .

Given $2x + 3y = \sin y$ . Differentiate both sides w.r.t. $x$ :

$\dfrac{d}{dx}(2x) + \dfrac{d}{dx}(3y) = \dfrac{d}{dx}(\sin y)$

$2 + 3\dfrac{dy}{dx} = \cos y\,\dfrac{dy}{dx}.$

Collect the $\dfrac{dy}{dx}$ terms: …

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CBSE 2024Set ANNUAL1 mark
$y = e^x + y^2$
›Reveal solutionSolution

This is an implicit relation; differentiate both sides with respect to x and collect the $\dfrac{dy}{dx}$ terms.

Given $y = e^x + y^2$ .

Differentiate both sides with respect to $x$ :

$\dfrac{dy}{dx} = e^x + 2y\dfrac{dy}{dx}$

Collect the $\dfrac{dy}{dx}$ terms on one side: …

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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 understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

The one key move: $y$ is really $y(x)$

The procedure

Worked example

Q.If 1−x2+1−y2=a(x−y)\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)1−x2​+1−y2​=a(x−y), prove that dydx=1−y21−x2\dfrac{dy}{dx} = \sqrt{\dfrac{1 - y^2}{1 - x^2}}dxdy​=1−x21−y2​​.

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

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