Q.If $y = \dfrac{ax+b}{cx+d}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{ab-cd}{(cx+d)^2}$ (B) $\dfrac{ax-c}{(cx+d)^2}$ (C) $\dfrac{ac-bd}{(cx+d)^2}$ (D) $\dfrac{ad-bc}{(cx+d)^2}$

Showing the 12 most recent of 62 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Derivative of $\cot x$ is

   (a) $\sec^2 x$

   (b) $-\csc^2 x$

   (c) $\csc^2 x$

   (d) $\csc x \cot x$

   ›Reveal solutionSolution

   This is a standard trigonometric derivative, derivable from $\cot x = \cos x/\sin x$ via the quotient rule.

   Write $\cot x = \dfrac{\cos x}{\sin x}$. By the quotient rule:

   $\dfrac{d}{dx}\left(\dfrac{\cos x}{\sin x}\right) = \dfrac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2x} = \dfrac{-\sin^2x-\cos^2x}{\sin^2x} = \dfrac{-1}{\sin^2x} = -\csc^2x$

   (using $\sin^2x+\cos^2x=1$).

   ✓Final answer

   (b) $-\csc^2x$.

2. CBSE 2026Set ANNUAL1 mark
   Q.$\frac{d}{dx}(x+a) =$ ______.

   ›Reveal solutionSolution

   $\frac{d}{dx}(x+a) = 1$.

   Using the sum rule of differentiation: $\frac{d}{dx}(x+a) = \frac{d}{dx}(x) + \frac{d}{dx}(a)$.

   The derivative of $x$ with respect to $x$ is $1$, and the derivative of a constant $a$ is $0$.

   So $\frac{d}{dx}(x+a) = 1 + 0 = 1$.

   ✓Final answer

   $\frac{d}{dx}(x+a) = 1$.

3. CBSE 2026Set ANNUAL1 mark
   Q.Find: d/dx (sin x cos x + 5x + 3)

   ›Reveal solutionSolution

   Rewrite sin x cos x as (1/2)sin 2x before differentiating, then differentiate term by term.

   Use the identity $\sin x\cos x = \dfrac12\sin 2x$:

   $f(x) = \frac12\sin2x + 5x + 3$

   Differentiate term by term:

   $f'(x) = \frac12\cdot2\cos2x + 5 + 0 = \cos2x + 5$

   (Equivalently, by the product rule: $\dfrac{d}{dx}(\sin x\cos x) = \cos x\cos x + \sin x(-\sin x) = \cos^2x-\sin^2x=\cos2x$, same result.)

   ✓Final answer

   $\dfrac{d}{dx}(\sin x\cos x+5x+3) = \cos2x+5$.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The derivative of $\tan x$ w.r.t. 'x' is:

   (a) $\sec x$

   (b) $\sec^3 x$

   (c) $\sec^2 x$

   (d) $-\sec x$

   ›Reveal solutionSolution

   Differentiating $\tan x = \dfrac{\sin x}{\cos x}$ using the quotient rule gives $\sec^2 x$.

   Write $\tan x = \dfrac{\sin x}{\cos x}$. By the quotient rule,

   $\dfrac{d}{dx}\left(\dfrac{\sin x}{\cos x}\right) = \dfrac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x} = \dfrac{\cos^2x+\sin^2x}{\cos^2x} = \dfrac{1}{\cos^2x} = \sec^2x$

   ✓Final answer

   $\dfrac{d}{dx}(\tan x) = \sec^2 x$, which is option (c).

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): The derivative of a constant function is zero. Reason (R): A constant function does not change, hence its rate of change is zero.

   (a) Both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are correct, but Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is correct, but Reason (R) is incorrect.

   (d) Assertion (A) is incorrect, but Reason (R) is correct.

   ›Reveal solutionSolution

   Both statements are true, and (R) correctly explains why (A) holds: a constant function's value never changes, so its derivative (rate of change) is zero.

   Assertion (A): For $f(x)=c$ (a constant), $f'(x) = \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h} = \lim_{h\to0}\dfrac{c-c}{h} = \lim_{h\to0} 0 = 0$. So (A) is TRUE.

   Reason (R): A constant function has the same value everywhere, i.e. it does not change as $x$ varies — so its rate of change is genuinely zero. This is precisely the reasoning that proves (A). So (R) is TRUE and is the correct explanation of (A).

   ✓Final answer

   Option (a): Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct explanation of Assertion (A).

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the columns — Column A: $\dfrac{d}{dx}\sin x$. Choose its correct equivalent from Column B.

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

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

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

   (d) $-\sin x$

   (e) $\sin x$

   (f) $-\cos x$

   (g) $\cos x$

   ›Reveal solutionSolution

   $\dfrac{d}{dx}\sin x = \cos x$, a standard derivative result.

   By first principles, $\dfrac{d}{dx}\sin x = \lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h}$. Using $\sin(x+h)-\sin x = 2\cos\left(x+\tfrac{h}{2}\right)\sin\tfrac{h}{2}$, this limit evaluates to $\cos x \cdot 1 = \cos x$ (since $\lim_{h\to0}\tfrac{\sin(h/2)}{h/2}=1$).

   ✓Final answer

   The correct match is (g) $\cos x$.

7. CBSE 2026Set ANNUAL1 mark
   Q.Write derivative of $\cos x$.

   ›Reveal solutionSolution

   $\dfrac{d}{dx}\cos x = -\sin x$, a standard derivative result.

   By first principles, $\dfrac{d}{dx}\cos x = \lim_{h\to0}\dfrac{\cos(x+h)-\cos x}{h}$. Using $\cos(x+h)-\cos x = -2\sin\left(x+\tfrac{h}{2}\right)\sin\tfrac{h}{2}$, the limit evaluates to $-\sin x \cdot 1 = -\sin x$.

   ✓Final answer

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

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.if $y = \log\left(\frac{e^x}{x^2}\right)$ then $\frac{dy}{dx}$ = ______.

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

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

   (c) $\frac{e - x}{e^x}$

   (d) $\frac{x - e}{ex}$

   ›Reveal solutionSolution

   Expand $\log\!\left(\dfrac{e^x}{x^2}\right)$ with log rules to get $x - 2\log x$, then differentiate; the answer is $\dfrac{x-2}{x}$ — option (ii).

   Simplify using $\log\dfrac{A}{B} = \log A - \log B$ and $\log e^x = x$:

   $y = \log\!\left(\frac{e^x}{x^2}\right) = \log e^x - \log x^2 = x - 2\log x.$

   Differentiate term by term, using $\dfrac{d}{dx}(\log x) = \dfrac{1}{x}$:

   $\frac{dy}{dx} = 1 - 2\cdot\frac{1}{x} = 1 - \frac{2}{x} = \frac{x - 2}{x}.$

   ✓Final answer

   $\dfrac{dy}{dx} = \dfrac{x-2}{x}$, which is option (ii).

9. CBSE 2026Set ANNUAL1 mark
   Q.If $y = e^{ax}$, then $x \cdot \frac{dy}{dx}$ = ______.

   ›Reveal solutionSolution

   $\dfrac{dy}{dx}=a e^{ax}$, so $x\cdot\dfrac{dy}{dx}=ax\,e^{ax}=axy$.

   Differentiate $y = e^{ax}$ using $\dfrac{d}{dx}(e^{ax}) = a e^{ax}$:

   $\frac{dy}{dx} = a e^{ax}.$

   Multiply by $x$:

   $x\cdot\frac{dy}{dx} = x\cdot a e^{ax} = ax\,e^{ax}.$

   Since $y = e^{ax}$, this can also be written compactly as $ax\,y$.

   ✓Final answer

   $x\cdot\dfrac{dy}{dx} = ax\,e^{ax} = axy$.

10. CBSE 2026Set MARCH1 markMCQ
    Q.If $y = ax + b$, $a$ and $b$ are constant then what will be $\dfrac{dy}{dx}$?

    (a) $a$

    (b) $b$

    (c) $a + b$

    (d) $0$

    ›Reveal solutionSolution

    $\dfrac{d}{dx}(ax+b)=a$.

    Using the sum and constant-multiple rules, $\dfrac{d}{dx}(ax)=a$ and $\dfrac{d}{dx}(b)=0$ (derivative of a constant). Hence

    $\frac{dy}{dx}=a.$

    ✓Final answer

    (a) $a$.

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\dfrac{d}{dx}(x^2 + 25x) =$

    (a) $2x$

    (b) 25

    (c) $2x + 25$

    (d) 0

    ›Reveal solutionSolution

    $\dfrac{d}{dx}(x^2+25x)=2x+25$.

    Using the power rule term-by-term: $\dfrac{d}{dx}(x^2)=2x$, and $\dfrac{d}{dx}(25x)=25$.

    Adding: $2x+25$.

    ✓Final answer

    (c) $2x+25$.

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\dfrac{d}{dx}(\tan x) =$

    (a) $\sec x$

    (b) $\cot x$

    (c) $\cot^2 x$

    (d) $\sec^2 x$

    ›Reveal solutionSolution

    $\dfrac{d}{dx}(\tan x)=\sec^2x$.

    Write $\tan x=\dfrac{\sin x}{\cos x}$ and apply the quotient rule:

    $\dfrac{d}{dx}\left(\dfrac{\sin x}{\cos x}\right)=\dfrac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^2x}=\dfrac{\cos^2x+\sin^2x}{\cos^2x}=\dfrac{1}{\cos^2x}=\sec^2x$.

    ✓Final answer

    (d) $\sec^2x$.