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

Showing the 12 most recent of 29 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 65/4/11 markMCQ
   Q.If $y = \sin^{-1} x$, then $(1 - x^2)\frac{d^2y}{dx^2}$ is equal to : (A) $x\frac{dy}{dx}$ (B) $-x\frac{dy}{dx}$ (C) $x^2\frac{dy}{dx}$ (D) $-x^2\frac{dy}{dx}$

   ›Reveal solutionSolution

   We find the first and second derivatives of $y = \sin^{-1} x$. By simplifying the first derivative before taking the second, we arrive at a differential equation that directly gives the value of $(1 - x^2)\frac{d^2y}{dx^2}$ as $x\frac{dy}{dx}$.

   The problem asks us to find the value of the expression $(1 - x^2)\frac{d^2y}{dx^2}$ given that $y = \sin^{-1} x$. This requires us to calculate both the first derivative ($\frac{dy}{dx}$) and the second derivative ($\frac{d^2y}{dx^2}$) of $y$ with respect to $x$. Once we have these, we will substitute them into the given expression and simplify.

   A key strategy in problems involving higher-order derivatives of inverse trigonometric functions is to simplify the first derivative expression before differentiating it again. This often involves eliminating square roots or fractions, which makes the subsequent differentiation much cleaner and less prone to errors.

   1. Find the first derivative, $\frac{dy}{dx}$.

      We are given the function $y = \sin^{-1} x$.

      The standard derivative of $\sin^{-1} x$ with respect to $x$ is:

      If $y = \sin^{-1} x$, then $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$.

      So, our first derivative is:

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

   2. Prepare for the second derivative by simplifying the first derivative expression.

      To make the calculation of the second derivative easier, we can rearrange the expression for $\frac{dy}{dx}$ to remove the square root from the denominator. This is a common and effective technique.

      Multiply both sides by $\sqrt{1 - x^2}$:

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

      Now, to eliminate the square root entirely, we square both sides of the equation:

      $\left(\sqrt{1 - x^2}\right)^2 \left(\frac{dy}{dx}\right)^2 = 1^2$

      $(1 - x^2) \left(\frac{dy}{dx}\right)^2 = 1$

      This form is much simpler to differentiate than the original fractional form.

   3. Find the second derivative, $\frac{d^2y}{dx^2}$.

      We will now differentiate the equation $(1 - x^2) \left(\frac{dy}{dx}\right)^2 = 1$ with respect to $x$. We need to apply the product rule on the left side and the chain rule for $\left(\frac{dy}{dx}\right)^2$.

      Let $u = (1 - x^2)$ and $v = \left(\frac{dy}{dx}\right)^2$.

      Then $\frac{du}{dx} = -2x$.

      And $\frac{dv}{dx} = 2\left(\frac{dy}{dx}\right) \frac{d}{dx}\left(\frac{dy}{dx}\right) = 2\frac{dy}{dx} \frac{d^2y}{dx^2}$.

      Applying the product rule, $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$: …

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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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