Q.If $x^y + y^x = a^b$ then prove that $\dfrac{dy}{dx} = -\left[ \dfrac{y\,x^{y-1} + y^x \log y}{x^y \log x + x\,y^{x-1}} \right]$.

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set 65/2/11 markMCQ
   Q.If $e^{-x} + e^{-y} = 2$, then $\frac{dy}{dx}$ is (A) $e^{x-y}$ (B) $e^{y-x}$ (C) $-e^{x-y}$ (D) $-e^{y-x}$

   ›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.

   1. 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:

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

   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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2. CBSE 2026Set A1 markMCQ
   Q.If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}}$ then $\frac{dy}{dx} =$

   (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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3. CBSE 2026Set A1 markMCQ
   Q.If $x^n + y^n = a^n$ then $\frac{dy}{dx} =$

   (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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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $2x + 3y = \sin y$, then $\frac{dy}{dx}$ is equal to

   (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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5. CBSE 2026Set ANNUAL1 mark
   Q.Find $\dfrac{dy}{dx}$ for the following : $2x+3y=\sin y$

   ›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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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $x^2+y^2=2$, then $\dfrac{dy}{dx}$ is equal to -

   (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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7. CBSE 2025Set ANNUAL1 mark
   Q.Find $\dfrac{dy}{dx}$, if $ax + by^2 = \cos y$. OR Find the integral $\displaystyle\int \dfrac{\log x}{x}\, dx$.

   ›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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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The value of dy/dx at (4, 1) of y³ − √x = 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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9. CBSE 2024Set D1 markMCQ
   Q.If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \text{ to } \infty}}}$ then $\frac{dy}{dx} =$

   (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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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.If $2x + 8y = \sin x$, then $\dfrac{dy}{dx}$ is:

    (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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11. CBSE 2024Set ANNUAL1 mark
    Q.Find $\dfrac{dy}{dx}$ for the following : $2x + 3y = \sin y$

    ›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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12. CBSE 2024Set ANNUAL1 mark
    Q.If $y = e^x + y^2$, then find $\dfrac{dy}{dx}$.

    ›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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