If $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$, prove that $\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}$.

If x = ecos 2t and y = esin 2t, prove that (dy)/(dx) = -(y log x)/(x log y).

NCERT Exemplar · Q49

x = e^{\cos 2t}

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Using implicit differentiation on the parametric equations $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$ , we find $\frac{dy}{dx} = -\frac{y \log x}{x \log y}$ by eliminating $t$ via logarithms and then differentiating.

We are given two parametric equations:

$x = e^{\cos 2t}$ and $y = e^{\sin 2t}$ .

We need to prove that $\frac{dy}{dx} = -\frac{y \log x}{x \log y}$ .

The direct approach — differentiating $x$ and $y$ with respect to $t$ and then using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ — is tempting. But notice the result involves $\log x$ and $\log y$ , which are not directly in the parametric forms. This suggests we should first eliminate the parameter $t$ by taking natural logarithms, then use implicit differentiation on the resulting relation between $x$ and $y$ .

Why implicit differentiation?

Implicit differentiation lets us find $\frac{dy}{dx}$ without solving for $y$ explicitly in terms of $x$ . Here, after taking logs, we get a simple relation: $\cos 2t = \log x$ and $\sin 2t = \log y$ . Squaring and adding gives $\log^2 x + \log^2 y = 1$ , which is an implicit equation linking $x$ and $y$ . Differentiating this directly yields the required derivative.

Let’s work through it step by step.

Take natural logarithms of both parametric equations.

From $x = e^{\cos 2t}$ , we have $\log x = \cos 2t$ .

From $y = e^{\sin 2t}$ , we have $\log y = \sin 2t$ .

(Here $\log$ denotes the natural logarithm, base $e$ .)
Eliminate $t$ by squaring and adding.

$\log^2 x + \log^2 y = \cos^2 2t + \sin^2 2t = 1$ .

So the relation between $x$ and $y$ is:

(\log x)^2 + (\log y)^2 = 1.

Differentiate both sides implicitly with respect to $x$ .

Remember that $y$ is a function of $x$ . Differentiate term by term:
- Derivative of $(\log x)^2$ : $2(\log x) \cdot \frac{1}{x}$ .
- Derivative of $(\log y)^2$ : $2(\log y) \cdot \frac{1}{y} \cdot \frac{dy}{dx}$ (by the chain rule).
- Derivative of the constant $1$ is $0$ .
So we get:

2 \frac{\log x}{x} + 2 \frac{\log y}{y} \cdot \frac{dy}{dx} = 0.

Solve for $\frac{dy}{dx}$ . Divide through by $2$ :

\frac{\log x}{x} + \frac{\log y}{y} \cdot \frac{dy}{dx} = 0.

Rearranging:

$\frac{\log y}{y} \cdot \frac{dy}{dx} = -\frac{\log x}{x}.$ …

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

Showing the 12 most recent of 27 on this concept.

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 $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$ , prove that $\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}$ .

Concept understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

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

The procedure

Worked example

Method: Eliminate the Parameter via Logarithms, Then Differentiate Implicitly

Steps

Common Mistakes

Q.If x=ecos⁡2tx = e^{\cos 2t}x=ecos2t and y=esin⁡2ty = e^{\sin 2t}y=esin2t, prove that dydx=−ylog⁡xxlog⁡y\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}dxdy​=−xlogyylogx​.

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Q.If $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$ , prove that $\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}$ .

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