Q.Find $\frac{dy}{dx}$ in the following: $\cos x^3 \cdot \sin^2 (x^5)$

Showing the 12 most recent of 115 on this concept.

1. CBSE 20191 mark
   Q.If $y = \cos (\sqrt{3x})$, then find $\frac{dy}{dx}$.

   ›Reveal solutionSolution

   Use the chain rule: differentiate the outer cosine function, then multiply by the derivative of the inner function $\sqrt{3x}$. The result is $\frac{dy}{dx} = -\frac{\sqrt{3}}{2\sqrt{x}} \sin(\sqrt{3x})$.

   The key idea here is the chain rule. Whenever you have a function of a function — like $\cos$ of something that itself depends on $x$ — you differentiate layer by layer. Think of it as peeling an onion: first the outer layer (cosine), then the next layer (the square root), and finally the innermost layer ($3x$). Each derivative multiplies together.

   Let’s walk through it step by step.

   1. Identify the composition. We have $y = \cos(u)$, where $u = \sqrt{3x}$. So $y$ depends on $u$, and $u$ depends on $x$. The chain rule says:

   $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$

   2. Differentiate the outer function. The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$. So:

   $\frac{dy}{du} = -\sin(u) = -\sin(\sqrt{3x}).$

   3. Differentiate the inner function $u = \sqrt{3x}$. Write $\sqrt{3x}$ as $(3x)^{1/2}$. Using the power rule and chain rule again (or directly):

   $\frac{du}{dx} = \frac{1}{2}(3x)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x}}.$

   Simplify: $\frac{3}{2\sqrt{3x}} = \frac{\sqrt{3}}{2\sqrt{x}}$, because $\frac{3}{\sqrt{3}} = \sqrt{3}$. …

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2. CBSE 2020Set 65/3/11 mark
   Q.Differentiate $\sin^2(\sqrt{x})$ with respect to $x$.

   ›Reveal solutionSolution

   We have a composition of three functions: squaring, sine, and square root. The chain rule peels them off one layer at a time, giving $\frac{d}{dx}\sin^2(\sqrt{x}) = \sin(\sqrt{x})\cos(\sqrt{x}) \cdot \frac{1}{\sqrt{x}}$.

   The chain rule is our tool for differentiating composite functions. When a function is built by nesting one operation inside another, we differentiate from the outside in, multiplying the derivative of each layer as we go.

   Here $\sin^2(\sqrt{x})$ is really $[\sin(\sqrt{x})]^2$, so we have three nested functions:

   • Outermost: squaring something
   • Middle: taking the sine of something
   • Innermost: taking the square root of $x$

   The chain rule says: differentiate the outer function (leaving the inside alone), then multiply by the derivative of what's inside, and repeat until you reach $x$.

   ---

   Step-by-step differentiation:

   1. Differentiate the outer square.

      If $u = \sin(\sqrt{x})$, then we're differentiating $u^2$. The power rule gives $2u$, so:

   $\frac{d}{dx}[\sin(\sqrt{x})]^2 = 2\sin(\sqrt{x}) \cdot \frac{d}{dx}[\sin(\sqrt{x})]$

   2. Differentiate the sine layer.

      Now we need $\frac{d}{dx}[\sin(\sqrt{x})]$. The derivative of $\sin(v)$ is $\cos(v)$, where $v = \sqrt{x}$:

   $\frac{d}{dx}[\sin(\sqrt{x})] = \cos(\sqrt{x}) \cdot \frac{d}{dx}[\sqrt{x}]$

   3. Differentiate the innermost square root.

      Finally, $\frac{d}{dx}[\sqrt{x}] = \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. …

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3. CBSE 2024Set 65/1/11 markMCQ
   Q.The derivative of $\sin(x^2)$ with respect to $x$ at $x = \sqrt{\pi}$ is : (A) $1$ (B) $-1$ (C) $-2\sqrt{\pi}$ (D) $2\sqrt{\pi}$

   ›Reveal solutionSolution

   The derivative of $\sin(x^2)$ is found using the Chain Rule: differentiate the outer sine, then multiply by the derivative of the inner $x^2$. At $x = \sqrt{\pi}$, the result is $-2\sqrt{\pi}$, which corresponds to option (C).

   The key to this problem is recognizing that $\sin(x^2)$ is a composite function. You have an outer function, $\sin(\text{something})$, and an inner function, $x^2$. When you need the derivative of a composition like this, the Chain Rule is your only reliable tool.

   Why does the Chain Rule work? Think of it as peeling an onion: you first differentiate the outer layer (sine) while keeping the inner layer untouched, then multiply by the derivative of the inner layer. This gives the rate of change of the whole expression with respect to $x$.

   Let’s walk through it step by step.

   1. Identify the outer and inner functions.

      Here, $f(u) = \sin(u)$ where $u = x^2$. So $f'(u) = \cos(u)$, and $u' = 2x$.

   2. Apply the Chain Rule.

      The derivative is:

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

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

   3. Evaluate at $x = \sqrt{\pi}$. Substitute $x = \sqrt{\pi}$ into the derivative:

   $2(\sqrt{\pi}) \cos\left( (\sqrt{\pi})^2 \right) = 2\sqrt{\pi} \cos(\pi).$

   4. Simplify $\cos(\pi)$. From the unit circle, $\cos(\pi) = -1$. So: $$$$

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4. 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 $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$: …

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5. 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, …

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6. 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 $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$: …

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7. 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, …

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8. 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 $y=\cos u$.

   …

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9. 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 $v=e^x$: …

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

    …

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11. 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: $\dfrac{d}{dx}e^{x^n}=nx^{n-1}e^{x^n}$.

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

    …

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12. 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 $\dfrac{du}{dx}$.

    …

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