Differentiate with respect to $x$: $\dfrac{x^2 \cos \frac{\pi}{4}}{\sin x}$.

Differentiate with respect to x: dfracx² cos (π)/(4)sin x.

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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. …

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The key idea is to treat the constant $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ as a factor, then apply the Quotient Rule to $\frac{x^2}{\sin x}$ . The derivative is $\frac{\sqrt{2}}{2} \cdot \frac{2x \sin x - x^2 \cos x}{\sin^2 x}$ .

We start by noticing that $\cos\frac{\pi}{4}$ is just a number — it does not depend on $x$ . That constant simplifies the problem immediately. Since $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ , the function becomes:

f(x) = \frac{\sqrt{2}}{2} \cdot \frac{x^2}{\sin x}

The factor $\frac{\sqrt{2}}{2}$ can be pulled out of the derivative (constant multiple rule). So the real work is differentiating $\frac{x^2}{\sin x}$ .

Why the Quotient Rule? Because we have one function of $x$ (namely $x^2$ ) divided by another function of $x$ (namely $\sin x$ ). The Quotient Rule says:

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

Here $u = x^2$ and $v = \sin x$ . Let’s apply it step by step.

Differentiate the numerator: $u' = \frac{d}{dx}(x^2) = 2x$ .
Differentiate the denominator: $v' = \frac{d}{dx}(\sin x) = \cos x$ .
Plug into the Quotient Rule:

\frac{d}{dx}\left(\frac{x^2}{\sin x}\right) = \frac{(2x)(\sin x) - (x^2)(\cos x)}{(\sin x)^2}

That’s the derivative of the inner fraction.

Now multiply by the constant $\frac{\sqrt{2}}{2}$ : $f'(x) = \frac{\sqrt{2}}{2} \cdot \frac{2x \sin x - x^2 \cos x}{\sin^2 x}$ …

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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: …

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CBSE 2026Set ANNUAL1 mark
$\frac{d}{dx}(x+a) =$
›Reveal solutionSolution

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

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

…

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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:

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

…

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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, …

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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.

…

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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.

…

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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 $\cos(x+h)-\cos x = -2\sin\left(x+\tfrac{h}{2}\right)\sin\tfrac{h}{2}$ , …

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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 $\log e^x = x$ :

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

…

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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 $x$ :

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

…

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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 cons …

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(x^2 + 25x) =$
(a) $2x$
(b) 25
(c) $2x + 25$
(d) 0
›Reveal solutionSolution

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

…

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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: …

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Q.Differentiate with respect to xxx: x2cos⁡π4sin⁡x\dfrac{x^2 \cos \frac{\pi}{4}}{\sin x}sinxx2cos4π​​.

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Q.Differentiate with respect to $x$ : $\dfrac{x^2 \cos \frac{\pi}{4}}{\sin x}$ .

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