Q.Find the second order derivative of the function: $x^{20}$

Showing the 12 most recent of 22 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\frac{d^2}{dx^2}(\sin 2x) =$

   (a) $4\sin 2x$

   (b) $4\cos^2 2x$

   (c) $-4\sin 2x$

   (d) $2\sin 4x$

   ›Reveal solutionSolution

   $\frac{d^2}{dx^2}(\sin 2x) = -4\sin 2x$.

   First derivative (chain rule):

   $\frac{d}{dx}(\sin 2x) = 2\cos 2x.$

   Second derivative:

   $\frac{d}{dx}(2\cos 2x) = 2\cdot(-\sin 2x)\cdot 2 = -4\sin 2x.$

   ✓Final answer

   (c) $-4\sin 2x$.

2. CBSE 2026Set ANNUAL1 mark
   Q.If y = 8e⁻³ˣ, find d²y/dx².

   ›Reveal solutionSolution

   Differentiate $y=8e^{-3x}$ twice using the chain rule.

   $\dfrac{dy}{dx} = 8\cdot(-3)e^{-3x} = -24e^{-3x}$

   $\dfrac{d^2y}{dx^2} = -24\cdot(-3)e^{-3x} = 72e^{-3x}$

   ✓Final answer

   $\dfrac{d^2y}{dx^2} = 72e^{-3x}$.

3. CBSE 2026Set ANNUAL1 mark
   Q.Find the second derivative for the function y=sin x + e^{2x}.

   ›Reveal solutionSolution

   $y''=-\sin x+4e^{2x}$.

   Concept. The second derivative is found by differentiating the first derivative; use $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}e^{kx}=k\,e^{kx}$.

   Steps.

   • $y=\sin x+e^{2x}$.

   • First derivative: $y'=\cos x+2e^{2x}$.

   • Second derivative: $y''=\dfrac{d}{dx}(\cos x)+\dfrac{d}{dx}(2e^{2x})=-\sin x+4e^{2x}$.

   ✓Final answer

   $\dfrac{d^2y}{dx^2}=-\sin x+4e^{2x}$.

4. CBSE 2025Set ANNUAL1 mark
   Q.Find the second order derivative of the function $y = \log x$.

   ›Reveal solutionSolution

   Differentiate $y=\log x$ twice.

   $y=\log x \implies \dfrac{dy}{dx}=\dfrac1x$

   Differentiating again: $\dfrac{d^2y}{dx^2}=-\dfrac{1}{x^2}$

   ✓Final answer

   $\dfrac{d^2y}{dx^2} = -\dfrac{1}{x^2}$

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $y = 2\sin x + 3\cos x$ then $\dfrac{d^2 y}{dx^2} =$

   (a) $y$

   (b) $\dfrac{1}{y}$

   (c) $-y$

   (d) $-\dfrac{1}{y}$

   ›Reveal solutionSolution

   Differentiate twice — the second derivative comes back around to -y, a classic SHM-type result.

   $y = 2\sin x + 3\cos x$

   $\dfrac{dy}{dx} = 2\cos x - 3\sin x$

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

   ✓Final answer

   (c) -y.

6. CBSE 2025Set ANNUAL1 mark
   Q.Find the second-order derivative of $x\cos x$ w.r.t. $x$.

   ›Reveal solutionSolution

   Apply the product rule twice.

   Let $y=x\cos x$.

   First derivative (product rule on $x$ and $\cos x$):

   $y'=\frac{d}{dx}(x)\cos x+x\frac{d}{dx}(\cos x)=1\cdot\cos x+x(-\sin x)=\cos x-x\sin x.$

   Second derivative (differentiate $\cos x$ and the product $x\sin x$):

   $y''=\frac{d}{dx}(\cos x)-\frac{d}{dx}(x\sin x)=-\sin x-\big(\sin x+x\cos x\big).$

   Simplifying,

   $y''=-\sin x-\sin x-x\cos x=-2\sin x-x\cos x.$

   ✓Final answer

   $\dfrac{d^2y}{dx^2}=-2\sin x-x\cos x$

7. CBSE 2024Set D1 markMCQ
   Q.If $y = x^{20}$ then $\frac{d^2y}{dx^2} =$

   (a) $x^{18}$

   (b) $20x^{19}$

   (c) $380x^{18}$

   (d) $x^{19}$

   ›Reveal solutionSolution

   $\frac{d^2y}{dx^2} = 380x^{18}$.

   Use the power rule $\frac{d}{dx}x^n = nx^{n-1}$ twice.

   First derivative:

   $\frac{dy}{dx} = 20x^{19}.$

   Second derivative:

   $\frac{d^2y}{dx^2} = 20 \times 19\, x^{18} = 380x^{18}.$

   ✓Final answer

   (C) $380x^{18}$.

8. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $y=\log x$ then $\frac{d^2y}{dx^2}=$

   (a) $-\frac{1}{x^2}$

   (b) $\frac{1}{x^2}$

   (c) $-\frac{1}{x}$

   (d) $\frac{1}{x}$

   ›Reveal solutionSolution

   Differentiate y=log x twice: first derivative is 1/x, second derivative is -1/x^2.

   $y=\log x \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x}=x^{-1}$

   $\dfrac{d^2y}{dx^2}=-x^{-2}=-\dfrac{1}{x^2}$

   ✓Final answer

   (a) $-\dfrac{1}{x^2}$.

9. CBSE 2024Set ANNUAL1 mark
   Q.Find the second-order derivative of $\log x$.

   ›Reveal solutionSolution

   Differentiate $\log x$ twice.

   Let $y = \log x$. The first derivative is

   $\dfrac{dy}{dx} = \dfrac{1}{x} = x^{-1}.$

   Differentiating again,

   $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(x^{-1}\right) = -x^{-2} = -\dfrac{1}{x^2}.$

   ✓Final answer

   $\boldsymbol{-\dfrac{1}{x^2}}$

10. CBSE 2023Set E1 markMCQ
    Q.$\dfrac{d^{2}}{dx^{2}}\left(e^{5x}\right) =$

    (a) $e^{5x}$

    (b) $10e^{5x}$

    (c) $5e^{5x}$

    (d) $25e^{5x}$

    ›Reveal solutionSolution

    $\dfrac{d}{dx}e^{5x}=5e^{5x}$, and differentiating again gives $25e^{5x}$.

    First derivative: $\dfrac{d}{dx}e^{5x}=5e^{5x}$.

    Second derivative: $\dfrac{d}{dx}\left(5e^{5x}\right)=5\cdot 5e^{5x}=25e^{5x}.$

    ✓Final answer

    (d) $25e^{5x}$.

11. CBSE 2023Set ANNUAL1 markMCQ
    Q.If $y=x \cdot \log_e x$, then the value of $\frac{d^2y}{dx^2}$ will be:

    (a) $\frac{1}{1+x}$

    (b) $\frac{1}{x}$

    (c) $\log_e(1+x)$

    (d) $1+\log_e x$

    ›Reveal solutionSolution

    Differentiate $y=x\log_e x$ twice using the product rule.

    Given $y=x\log_e x$.

    First derivative (product rule, $u=x$, $v=\log_e x$):

    $\dfrac{dy}{dx}=1\cdot\log_e x + x\cdot\dfrac{1}{x}=\log_e x+1$

    Second derivative:

    $\dfrac{d^2y}{dx^2}=\dfrac{1}{x}+0=\dfrac{1}{x}$

    ✓Final answer

    (b) $\dfrac{1}{x}$.

12. CBSE 2022Set ANNUAL1 markMCQ
    Q.$\dfrac{d^2}{dx^2}\sin(3x+5) = ?$

    (a) $\sin(3x+5)$

    (b) $9\cos(3x+5)$

    (c) $-9\sin(3x+5)$

    (d) $9\sin(3x+5)$

    ›Reveal solutionSolution

    Differentiate twice by the chain rule; each derivative brings down a factor of 3, and differentiating sine twice turns it into $-\sin$.

    Let $y=\sin(3x+5)$.

    First derivative: $\dfrac{dy}{dx}=3\cos(3x+5)$.

    Second derivative: $\dfrac{d^2y}{dx^2}=3\cdot\big(-\sin(3x+5)\big)\cdot3=-9\sin(3x+5)$.

    ✓Final answer

    (c) $-9\sin(3x+5)$.