Select the appropriate option from the given alternative. If $y = \dfrac{x-4}{\sqrt{x}+2}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{1}{x+4}$ (B) $\dfrac{\sqrt{x}}{(\sqrt{x}+2)^2}$ (C) $\dfrac{1}{2\sqrt{x}}$ (D) $\dfrac{x}{(\sqrt{x}+2)^2}$

Select the appropriate option from the given alternative. If y = dfracx-4√(x)+2, then (dy)/(dx) = (A) (1)/(x+4

MISCELLANEOUS EXERCISE 9 (I) · Q58

y = \dfrac{x-4}{\sqrt{x}+2}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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✓ Free question

Concept understanding — Rules of Differentiation

Computing every derivative directly from the limit definition ("from first principle") is correct but impractical for anything beyond the simplest functions. The five rules below, proved once from the definition, make differentiation a mechanical, algebraic process for any combination of known derivatives — no further limit computation is needed once they are established. Throughout, $u=u(x)$ , $v=v(x)$ are differentiable functions of $x$ , and $k$ is a constant.

Sum Rule (Theorem 10.2). $\dfrac{d}{dx}(u+v)=\dfrac{du}{dx}+\dfrac{dv}{dx}$ — differentiate term by term. Extends to any finite sum.

Product Rule (Theorem 10.3). $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ , i.e. $(uv)'=uv'+vu'$ — "first times derivative of second, plus second times derivative of first." Extends to three or more factors, e.g. $(uvw)'=u'vw+uv'w+uvw'$ : differentiate one factor at a time, summing over which factor is differentiated.

Quotient Rule (Theorem 10.4). For $v\ne0$ , $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\,u'-u\,v'}{v^2}$ — "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared."

Chain Rule (Theorem 10.5). If $y=f(u)$ and $u=g(x)$ are both differentiable, so $y=f(g(x))$ , then

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = f'(g(x))\,g'(x)

— differentiate the outer function $f$ with respect to its argument $u=g(x)$ (the inner function), then multiply by the derivative of the inner function. This is the single most-used rule in the chapter: every composite function (a power of an expression, $\sin$ of an expression, $e^{\text{expression}}$ , and so on) needs it.

Constant Multiple Rule (Theorem 10.6). $\dfrac{d}{dx}\big[k\,f(x)\big] = k\,f'(x)$ — a constant factor simply carries through differentiation unchanged.

All five are proved the same way: express the difference quotient of the combined function algebraically in terms of the difference quotients of $u$ and $v$ (inserting-and-subtracting a convenient cross term for the product and quotient rules; splitting into $\Delta y/\Delta u \times \Delta u/\Delta x$ for the chain rule), then take the limit of each piece separately, using that a differentiable function is continuous (Theorem 10.1) wherever the proof needs $v(x+\Delta x)\to v(x)$ or $\Delta u\to0$ .

Tip

These five rules, together with the derivative table of standard functions, are jointly sufficient for every differentiation problem in this chapter — no problem ever requires returning to the first-principle limit once the rules and the table are available.

$y=\dfrac{x-4}{\sqrt x+2}$ . With $u=x-4,\ u'=1$ and $v=\sqrt x+2,\ v'=\dfrac1{2\sqrt x}$ :

$\dfrac{dy}{dx}=\dfrac{(\sqrt x+2)(1)-(x-4)\cdot\frac1{2\sqrt x}}{(\sqrt x+2)^2}$

Combine the numerator over $2\sqrt x$ : $\dfrac{2\sqrt x(\sqrt x+2)-(x-4)}{2\sqrt x}=\dfrac{2x+4\sqrt x-x+4}{2\sqrt x}=\dfrac{x+4\sqrt x+4}{2\sqrt x}=\dfrac{(\sqrt x+2)^2}{2\sqrt x}$

(since $x+4\sqrt x+4=(\sqrt x)^2+4\sqrt x+4=(\sqrt x+2)^2$ ).

$\dfrac{dy}{dx}=\dfrac{(\sqrt x+2)^2/(2\sqrt x)}{(\sqrt x+2)^2}=\dfrac1{2\sqrt x}$

✓Final answer

\dfrac1{2\sqrt x}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 62 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\cot x$
(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
$-\csc^2x$
CBSE 2026Set ANNUAL1 mark
$\frac{d}{dx}(x+a) =$
›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
$\frac{d}{dx}(x+a) = 1$
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
$\dfrac{d}{dx}(\sin x\cos x+5x+3) = \cos2x+5$
CBSE 2026Set ANNUAL1 markMCQ
$\tan x$
(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
$\dfrac{d}{dx}(\tan x) = \sec^2 x$
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
$Option (a): Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct explanation of Assertion (A).$
CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx}\sin x$
(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
$\cos x$
CBSE 2026Set ANNUAL1 mark
$\cos x$
›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
$\dfrac{d}{dx}(\cos x) = -\sin x$
CBSE 2026Set ANNUAL1 markMCQ
$y = \log\left(\frac{e^x}{x^2}\right)$
(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
$\dfrac{dy}{dx} = \dfrac{x-2}{x}$
CBSE 2026Set ANNUAL1 mark
$y = e^{ax}$
›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
$x\cdot\dfrac{dy}{dx} = ax\,e^{ax} = axy$
CBSE 2026Set MARCH1 markMCQ
$y = ax + b$
(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
$a$
CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(x^2 + 25x) =$
(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
$2x+25$
CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(\tan x) =$
(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
$\sec^2x$

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