Q.The equation of the normal to the parabola $y^2 = 8x$ at its origin is ________.

Showing the 12 most recent of 42 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.The edge of a variable cube is increasing at the rate of 3 cm/s. The volume of the cube is increasing at the rate of __________ while the edge is 10 cm long.

   ›Reveal solutionSolution

   Use $V=e^3$ and the chain rule $dV/dt = 3e^2\,de/dt$.

   Let $e$ be the edge; $V=e^3$, so $\dfrac{dV}{dt}=3e^2\dfrac{de}{dt}$.

   Given $\dfrac{de}{dt}=3$ cm/s and $e=10$ cm:

   …

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6. CBSE 2026Set ANNUAL1 mark
   Q.The radius of an air bubble is increasing at the rate of $1/2$ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

   ›Reveal solutionSolution

   Use $V=\dfrac43\pi r^3$ and $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.

   Given $\dfrac{dr}{dt}=\dfrac12$ cm/s, at $r=1$ cm:

   …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.Radius of a circle is increasing at the rate of 1/π m/s. Rate of change of its circumference is:

   (a) 4π m/s

   (b) 2 m/s

   (c) 2π m/s

   (d) 4 m/s

   ›Reveal solutionSolution

   Since $C = 2\pi r$, differentiating both sides w.r.t. time gives $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$ directly.

   The circumference of a circle of radius $r$ is $C = 2\pi r$.

   Differentiating with respect to time $t$:

   $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$

   Given $\frac{dr}{dt} = \frac{1}{\pi}$ m/s, substitute: …

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8. 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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9. CBSE 2025Set 65/2/11 markMCQ
   Q.A cylindrical tank of radius 10 cm is being filled with sugar at the rate of $100\pi$ cm$^3$/s. The rate at which the height of the sugar inside the tank is increasing is: (A) $0.1$ cm/s (B) $0.5$ cm/s (C) $1$ cm/s (D) $1.1$ cm/s

   ›Reveal solutionSolution

   The volume of a cylinder is $V = \pi r^2 h$. Since the radius is constant, the rate of change of volume with respect to time is $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$. Given $\frac{dV}{dt} = 100\pi$ cm³/s and $r = 10$ cm, solving gives $\frac{dh}{dt} = 1$ cm/s. The correct option is (C).

   This is a classic Related Rates problem. The core idea is that when two quantities are linked by a geometric formula (here, volume and height of a cylinder), their rates of change with respect to time are also linked. You differentiate the relationship with respect to time, plug in what you know, and solve for the unknown rate.

   The key insight: the tank’s radius is fixed at 10 cm. So as sugar pours in, the height increases, but the cross-sectional area stays the same. That means the volume increases at a constant rate per unit height — specifically, each 1 cm rise in height adds $\pi (10)^2 = 100\pi$ cm³ of volume. Since sugar is being added at exactly $100\pi$ cm³/s, the height must be rising at 1 cm/s.

   Let’s work it out formally.

   1. Write the relationship between volume and height. For a cylinder, $V = \pi r^2 h$. Here $r = 10$ cm, so

   $V = \pi (10)^2 h = 100\pi h.$

   2. Differentiate both sides with respect to time $t$.

      Since $r$ is constant, $\frac{dV}{dt} = 100\pi \frac{dh}{dt}$.

      This is the related rates equation — it tells us how fast the volume changes in terms of how fast the height changes.

   3. Substitute the given rate.

      We know $\frac{dV}{dt} = 100\pi$ cm³/s. So:

   $100\pi = 100\pi \frac{dh}{dt}.$

   4. Solve for $\frac{dh}{dt}$. Divide both sides by $100\pi$: …

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10. 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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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Radius of a circle is increasing at the rate of 2 m/s. Rate of change of its circumference is:

    (a) 4π m/s

    (b) 2 m/s

    (c) 2π m/s

    (d) 4 m/s

    ›Reveal solutionSolution

    Differentiate the circumference formula $C=2\pi r$ with respect to time and plug in $\dfrac{dr}{dt}$.

    Given $\dfrac{dr}{dt} = 2$ m/s. Circumference $C = 2\pi r$.

    Differentiating both sides with respect to time $t$: …

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12. 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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