Q.Show that the tangent at $P(x_1, y_1)$ on the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is $y\,y_1^{-1/2} + x\,x_1^{-1/2} = a^{1/2}$.

Showing the 12 most recent of 103 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.If $\sin^{-1} x = y$, then $\frac{dy}{dx}$ is: (A) $\cos^{-1} x$ (B) $\cos y$ (C) $\frac{1}{1-x^2}$ (D) $\sec y$

   ›Reveal solutionSolution

   The derivative of $\sin^{-1} x$ is found by implicit differentiation of $x = \sin y$, giving $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}$, which matches option (B) $\cos y$ only if we interpret it as $\sec y$ — but careful: the correct form is $\frac{1}{\sqrt{1-x^2}}$, and among the given choices, (B) $\cos y$ is actually $\frac{1}{\cos y}$? No — let's check: $\cos y = \sqrt{1-x^2}$, so $\frac{1}{\cos y} = \sec y$, which is option (D). The final answer is (D) $\sec y$.

   The core idea: when you have an inverse trigonometric function, the easiest way to differentiate it is to rewrite it as a direct trigonometric equation and then use implicit differentiation. This avoids memorising a dozen formulas and builds from what you already know — the derivative of $\sin$ and the chain rule.

   Let $\sin^{-1} x = y$. This means $x = \sin y$, and importantly, $y$ is restricted to $[-\pi/2, \pi/2]$ so that $\cos y \ge 0$.

   1. Start with the relation:

   $x = \sin y$

   2. Differentiate both sides with respect to $x$. Remember $y$ is a function of $x$, so we use the chain rule on the right:

   $\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$

   $1 = \cos y \cdot \frac{dy}{dx}$

   3. Solve for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{1}{\cos y}$

   4. Now, $\cos y$ can be expressed in terms of $x$. Since $\sin y = x$, we use the identity $\sin^2 y + \cos^2 y = 1$:

   $\cos^2 y = 1 - \sin^2 y = 1 - x^2$

   $\cos y = \sqrt{1 - x^2} \quad (\text{positive because } y \in [-\pi/2, \pi/2])$

   5. Therefore: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$ …

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2. CBSE 2026Set V11 markMCQ
   Q.If $x - y = \pi$ then $\frac{dy}{dx}$

   (a) $\pi$

   (b) $-\pi$

   (c) $1$

   (d) $-1$

   ›Reveal solutionSolution

   Differentiating the constant-difference relation gives $\dfrac{dy}{dx}=1$; answer (c).

   Differentiate $x-y=\pi$ (a constant) with respect to $x$: …

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

   (a) $\frac{1}{x^n}$

   (b) $n$

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

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

   ›Reveal solutionSolution

   $\frac{d}{dx}(\log x^n) = \frac{n}{x}$.

   First simplify with the log power rule:

   $\log x^n = n\log x.$

   Differentiate: …

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4. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}(e^{x-a}) =$

   (a) $e^{x-a}$

   (b) $(x-a)e^{x-a}$

   (c) $e^x$

   (d) $-e^{x-a}$

   ›Reveal solutionSolution

   $\frac{d}{dx}\left(e^{x-a}\right) = e^{x-a}$.

   Here $a$ is a constant. Let $u = x-a$, so $\frac{du}{dx} = 1$.

   By the chain rule, …

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5. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}(\tan^{-1}\sqrt{x} + \cot^{-1}\sqrt{x}) =$

   (a) $\frac{\pi}{2}$

   (b) $0$

   (c) $1$

   (d) $\pi$

   ›Reveal solutionSolution

   The derivative is $0$ because the sum is a constant.

   For any real argument $t$, $\tan^{-1}t + \cot^{-1}t = \frac{\pi}{2}$. Taking $t=\sqrt{x}$,

   $\tan^{-1}\sqrt{x} + \cot^{-1}\sqrt{x} = \frac{\pi}{2}.$

   …

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

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

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

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

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

   ›Reveal solutionSolution

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

   Using the standard derivative $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ and the constant multiple rule: …

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7. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}\left\{\lim_{x\to 0}\frac{x^5-a^5}{x-a}\right\} =$

   (a) $a$

   (b) $0$

   (c) $5a^4$

   (d) $5$

   ›Reveal solutionSolution

   The inner limit is a constant, so the derivative is $0$.

   Evaluate the limit first (it does not depend on $x$ after the limit is taken):

   $\lim_{x\to a}\frac{x^5-a^5}{x-a} = 5a^4$ …

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

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

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

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

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

   ›Reveal solutionSolution

   $\frac{d}{dx}\cot^{-1}x = \frac{-1}{1+x^2}$.

   This is a standard result. From $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$, differentiating gives …

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9. CBSE 2026Set A1 markMCQ
   Q.$\frac{d}{dx}\sin^{-1}(3x - 4x^3) =$

   (a) $\frac{3}{\sqrt{1-x^2}}$

   (b) $\frac{-3}{\sqrt{1-x^2}}$

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

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

   ›Reveal solutionSolution

   $\frac{d}{dx}\sin^{-1}(3x-4x^3) = \frac{3}{\sqrt{1-x^2}}$.

   Use the identity (for the principal branch, $-\frac{1}{2}\le x\le\frac{1}{2}$):

   $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x,$

   which follows from $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ with $x=\sin\theta$.

   Differentiate: …

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

    (a) $4\cos 8x$

    (b) $4\cos 8x - \cos 2x$

    (c) $\cos 2x$

    (d) $\cos 2x - 4\cos 8x$

    ›Reveal solutionSolution

    Convert the product to a sum, then differentiate: $\frac{d}{dx}(\sin 3x\cos 5x)=4\cos 8x-\cos 2x$.

    Use the product-to-sum identity $\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$.

    $\sin 3x\cos 5x=\tfrac12[\sin 8x+\sin(-2x)]=\tfrac12[\sin 8x-\sin 2x]$.

    Differentiate term by term:

    …

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11. CBSE 2026Set ANNUAL1 mark
    Q.If $y = \frac{e^x}{\sin x}$, then find $\frac{dy}{dx}$.

    ›Reveal solutionSolution

    Apply the quotient rule to $y=e^x/\sin x$.

    $\dfrac{dy}{dx} = \dfrac{e^x\sin x - e^x\cos x}{\sin^2 x} = \dfrac{e^x(\sin x-\cos x)}{\sin^2x}$.

    …

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

    (a) $a^x \log a$

    (b) $\log a$

    (c) $a^x$

    (d) None of these

    ›Reveal solutionSolution

    $\dfrac{d}{dx}a^x = a^x \ln a$, using $a^x = e^{x\ln a}$.

    …

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