Q.Differentiate $y=\sin(3x^2+2x)$ with respect to $x$.

Showing the 12 most recent of 115 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}\sqrt{x^2+ax+1} =$

   (a) $\frac{x+a}{2\sqrt{x^2+ax+1}}$

   (b) $\frac{2x+a}{2\sqrt{x^2+ax+1}}$

   (c) $\frac{2x+a}{\sqrt{x^2+ax+1}}$

   (d) $\frac{1}{2\sqrt{x^2+ax+1}}$

   ›Reveal solutionSolution

   $\frac{d}{dx}\sqrt{x^2+ax+1} = \frac{2x+a}{2\sqrt{x^2+ax+1}}$.

   Let $u = x^2+ax+1$, so $\frac{du}{dx} = 2x+a$.

   Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$:

   $\frac{d}{dx}\sqrt{x^2+ax+1} = \frac{2x+a}{2\sqrt{x^2+ax+1}}.$

   ✓Final answer

   (b) $\frac{2x+a}{2\sqrt{x^2+ax+1}}$.

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

   (a) $2x\cos x^2$

   (b) $\cos x^2$

   (c) $x^2\cos x^2$

   (d) $x\cos x^2$

   ›Reveal solutionSolution

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

   Let $u = x^2$, so $\frac{du}{dx} = 2x$.

   By the chain rule,

   $\frac{d}{dx}\sin u = \cos u\cdot\frac{du}{dx} = \cos(x^2)\cdot 2x = 2x\cos(x^2).$

   ✓Final answer

   (a) $2x\cos x^2$.

3. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}\sqrt{\cot x} =$

   (a) $\frac{1}{2\sqrt{\cot x}}$

   (b) $\sqrt{\csc^2 x}$

   (c) $\frac{-\csc^2 x}{2\sqrt{\cot x}}$

   (d) $\frac{\csc^2 x}{2\sqrt{\cot x}}$

   ›Reveal solutionSolution

   $\frac{d}{dx}\sqrt{\cot x} = \frac{-\csc^2 x}{2\sqrt{\cot x}}$.

   Let $u = \cot x$, so $\frac{du}{dx} = -\csc^2 x$.

   Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$:

   $\frac{d}{dx}\sqrt{\cot x} = \frac{1}{2\sqrt{\cot x}}\cdot(-\csc^2 x) = \frac{-\csc^2 x}{2\sqrt{\cot x}}.$

   ✓Final answer

   (c) $\frac{-\csc^2 x}{2\sqrt{\cot x}}$.

4. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}(\cos\sqrt{x}) =$

   (a) $\sin\sqrt{x}$

   (b) $\frac{-\sin\sqrt{x}}{2\sqrt{x}}$

   (c) $\frac{\sin\sqrt{x}}{2\sqrt{x}}$

   (d) $\frac{1}{2\sqrt{x}}$

   ›Reveal solutionSolution

   $\frac{d}{dx}\cos\sqrt{x} = \frac{-\sin\sqrt{x}}{2\sqrt{x}}$.

   Let $u = \sqrt{x}$, so $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$.

   By the chain rule,

   $\frac{d}{dx}\cos u = -\sin u\cdot\frac{du}{dx} = -\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}} = \frac{-\sin\sqrt{x}}{2\sqrt{x}}.$

   ✓Final answer

   (b) $\frac{-\sin\sqrt{x}}{2\sqrt{x}}$.

5. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}(\cos x^3) =$

   (a) $-3x^2\sin x^3$

   (b) $\sin x^3$

   (c) $3x^2\sin x^3$

   (d) $3x^2$

   ›Reveal solutionSolution

   Chain rule on $\cos(x^3)$ gives $-3x^2\sin x^3$.

   Let the inner function be $u=x^3$, so $y=\cos u$.

   $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=(-\sin u)(3x^2)=-3x^2\sin x^3$.

   ✓Final answer

   (A) $-3x^2\sin x^3$.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the derivative of $\log(\cos e^x)$.

   (a) $-\tan e^x$

   (b) $e^x\tan e^x$

   (c) $-e^x\tan e^x$

   (d) $\tan e^x$

   ›Reveal solutionSolution

   Using the chain rule twice, $\dfrac{d}{dx}\log(\cos e^x)=-e^x\tan e^x$.

   Let $y=\log(\cos e^x)$. We differentiate using the chain rule, working from the outside in.

   Step 1: Differentiate $\log(u)$ where $u=\cos e^x$:

   $\frac{dy}{dx}=\frac{1}{\cos e^x}\cdot\frac{d}{dx}(\cos e^x)$

   Step 2: Differentiate $\cos(v)$ where $v=e^x$:

   $\frac{d}{dx}(\cos e^x)=-\sin(e^x)\cdot\frac{d}{dx}(e^x)=-\sin(e^x)\cdot e^x$

   Step 3: Combine:

   $\frac{dy}{dx}=\frac{-e^x\sin e^x}{\cos e^x}=-e^x\tan e^x$

   ✓Final answer

   The correct option is (c) $-e^x\tan e^x$.

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, $0 < x < 1$, then $\frac{dy}{dx}$ is equal to

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

   (b) $\frac{2}{4+x}$

   (c) $\frac{2}{1+x^2}$

   (d) $\frac{2}{x+x^2}$

   ›Reveal solutionSolution

   Put $x=\tan\phi$ so the expression simplifies to $\cos^{-1}(\cos2\phi)=2\phi=2\tan^{-1}x$, whose derivative is standard.

   Let $x=\tan\phi$. Then $\dfrac{1-x^2}{1+x^2} = \dfrac{1-\tan^2\phi}{1+\tan^2\phi} = \cos2\phi$.

   So $y=\cos^{-1}(\cos2\phi)=2\phi=2\tan^{-1}x$ (valid for $0<x<1$, i.e. $0<\phi<\pi/4$, so $2\phi$ is in the principal range).

   $\dfrac{dy}{dx} = \dfrac{2}{1+x^2}$.

   ✓Final answer

   The correct option is (c) $\dfrac{2}{1+x^2}$.

8. CBSE 2026Set ANNUAL1 mark
   Q.If $y = e^x + e^{x^2} + \ldots + e^{x^5}$, then find $\frac{dy}{dx}$.

   ›Reveal solutionSolution

   Differentiate each term $e^{x^n}$ using the chain rule: $\dfrac{d}{dx}e^{x^n}=nx^{n-1}e^{x^n}$.

   $y=e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}$

   $\dfrac{dy}{dx}=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$.

   ✓Final answer

   $\dfrac{dy}{dx}=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$.

9. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\dfrac{d}{dx}(\cos 3x) =$

   (a) $\sin 3x$

   (b) $-3\sin 3x$

   (c) $\cos 3x$

   (d) $-3\cos 3x$

   ›Reveal solutionSolution

   Differentiate $\cos(3x)$ using the chain rule: derivative of $\cos u$ is $-\sin u$, times $\dfrac{du}{dx}$.

   $\dfrac{d}{dx}(\cos 3x) = -\sin(3x) \cdot \dfrac{d}{dx}(3x) = -\sin(3x)\cdot 3 = -3\sin 3x$.

   ✓Final answer

   (b) $-3\sin 3x$.

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.$\dfrac{d}{dx} \tan^{-1}(x^2) =$

    (a) $\dfrac{2x}{1+x^4}$

    (b) $\dfrac{x}{1+x^2}$

    (c) $\dfrac{x^3}{1+x^2}$

    (d) None of these

    ›Reveal solutionSolution

    Use $\dfrac{d}{dx}\tan^{-1}u = \dfrac{1}{1+u^2}\cdot\dfrac{du}{dx}$ with $u=x^2$.

    $\dfrac{d}{dx}\tan^{-1}(x^2) = \dfrac{1}{1+(x^2)^2}\cdot \dfrac{d}{dx}(x^2) = \dfrac{2x}{1+x^4}$.

    ✓Final answer

    (a) $\dfrac{2x}{1+x^4}$.

11. CBSE 2026Set ANNUAL1 markMCQ
    Q.$\dfrac{d}{dx} e^{\sin x} =$

    (a) $e^{\sin x} \cdot \cos x$

    (b) $e^{\sin x}$

    (c) $\cos x$

    (d) None of these

    ›Reveal solutionSolution

    Differentiate $e^{u}$ with $u=\sin x$: derivative is $e^u \cdot \dfrac{du}{dx}$.

    $\dfrac{d}{dx}e^{\sin x} = e^{\sin x} \cdot \dfrac{d}{dx}(\sin x) = e^{\sin x}\cos x$.

    ✓Final answer

    (a) $e^{\sin x}\cdot\cos x$.

12. CBSE 2026Set ANNUAL1 markMCQ
    Q.$\dfrac{d}{dx}(\log \sec x) =$

    (a) $\tan x$

    (b) $\cot x$

    (c) $\csc x$

    (d) None of these

    ›Reveal solutionSolution

    Differentiate $\log(\sec x)$ using $\dfrac{d}{dx}\log u = \dfrac{u'}{u}$ with $u=\sec x$.

    $\dfrac{d}{dx}\log(\sec x) = \dfrac{1}{\sec x}\cdot \dfrac{d}{dx}(\sec x) = \dfrac{1}{\sec x}\cdot \sec x \tan x = \tan x$.

    ✓Final answer

    (a) $\tan x$.