If $f(x)=\dfrac{1}{x^n}$, for $x\ne 0$, $n\in N$, then prove that $f'(x)=-\dfrac{n}{x^{n+1}}$.

If f(x)=(1)/(xⁿ), for x≠ 0, n∈ N, then prove that f'(x)=-dfracnxⁿ⁺¹.

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f(x)=\dfrac{1}{x^n}

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✓ Free question

Concept understanding — Derivatives of Standard Functions

Applying the first-principle limit once to each basic elementary function builds a permanent table; every later problem then differentiates by combining this table with the rules of differentiation (sum/product/quotient/chain/constant-multiple), with no further limit ever required.

Algebraic functions.

\frac{d}{dx}(k)=0\ (k\text{ constant}), \qquad \frac{d}{dx}\!\left(x^n\right) = nx^{n-1}\ \text{for any real }n\ \text{(Corollaries 10.1–10.2 extend the integer case to rational, then any real, exponent).}

Logarithmic and exponential functions.

\frac{d}{dx}(\log x) = \frac1x,\qquad \frac{d}{dx}(\log_a x)=\frac{1}{x\log a},\qquad \frac{d}{dx}\!\left(a^x\right) = a^x\log a,\qquad \frac{d}{dx}\!\left(e^x\right)=e^x.

( $e^x$ is the unique elementary function that is its own derivative — the special case $a=e$ , $\log e=1$ .)

The six trigonometric functions.

\frac{d}{dx}(\sin x)=\cos x,\qquad \frac{d}{dx}(\cos x)=-\sin x,\qquad \frac{d}{dx}(\tan x)=\sec^2x,

\frac{d}{dx}(\sec x)=\sec x\tan x,\qquad \frac{d}{dx}(\operatorname{cosec}x)=-\operatorname{cosec}x\cot x,\qquad \frac{d}{dx}(\cot x)=-\operatorname{cosec}^2x.

Only $\sin x\to\cos x$ is derived directly from the limit definition (via the sum-to-product identity and $\lim_{\theta\to0}\sin\theta/\theta=1$ ); every other trig derivative follows from it using the chain rule ( $\cos x=\sin(\tfrac\pi2+x)$ ) or the quotient rule ( $\tan x=\sin x/\cos x$ , etc.) — so the whole trig table rests on a single limit.

The six inverse trigonometric functions (each domain-restricted to its principal branch):

\frac{d}{dx}\!\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\!\left(\cos^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}},\qquad \frac{d}{dx}\!\left(\tan^{-1}x\right)=\frac{1}{1+x^2},

\frac{d}{dx}\!\left(\cot^{-1}x\right)=-\frac{1}{1+x^2},\qquad \frac{d}{dx}\!\left(\sec^{-1}x\right)=\frac{1}{|x|\sqrt{x^2-1}},\qquad \frac{d}{dx}\!\left(\operatorname{cosec}^{-1}x\right)=-\frac{1}{|x|\sqrt{x^2-1}}.

$\sin^{-1}x$ and $\tan^{-1}x$ are derived by writing $x=\sin y$ (resp. $x=\tan y$ ) and differentiating implicitly (§10.4.3's technique, used here before that section is formally reached); the three complementary pairs ( $\cos^{-1}\leftrightarrow\sin^{-1}$ , $\cot^{-1}\leftrightarrow\tan^{-1}$ , and $\operatorname{cosec}^{-1}\leftrightarrow\sec^{-1}$ ) each follow from a co-function identity such as $\sin^{-1}x+\cos^{-1}x=\tfrac\pi2$ (a constant, so the two derivatives are exact negatives of each other).

Tip

Nearly every problem asking to "find the derivative" of an explicit function combines two or more of these table entries via the chain, product, or quotient rule — recognising which table entries are in play (is the outer layer a power? a trig function? a log?) is the whole skill being practised in Exercises 10.2 and 10.3.

Let $f(x)=\dfrac1{x^n}$ , $x\ne0,\ n\in N$ .

$f(x+h)-f(x)=\dfrac1{(x+h)^n}-\dfrac1{x^n}=\dfrac{x^n-(x+h)^n}{x^n(x+h)^n}$ .

By the binomial expansion, $(x+h)^n=x^n+nx^{n-1}h+\binom n2x^{n-2}h^2+\cdots$ , so $x^n-(x+h)^n=-nx^{n-1}h-\binom n2x^{n-2}h^2-\cdots=-h\!\left[nx^{n-1}+\binom n2x^{n-2}h+\cdots\right]$ .

$\dfrac{f(x+h)-f(x)}h=\dfrac{-\left[nx^{n-1}+\binom n2x^{n-2}h+\cdots\right]}{x^n(x+h)^n}$ .

As $h\to0$ : every term but $nx^{n-1}$ in the bracket vanishes, and $(x+h)^n\to x^n$ , giving $\dfrac{-nx^{n-1}}{x^n\cdot x^n}=-\dfrac n{x^{n+1}}$ .

✓Final answer

f'(x)=-\dfrac n{x^{n+1}}

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

CBSE 2026Set ANNUAL1 markMCQ
$y = mx + c$
(a) 3
(b) 1
(c) -3
(d) 2
›Reveal solutionSolution

With $f(x)=mx+c$ , $f(0)=c=1$ and $f'(0)=m=1$ , so $f(x)=x+1$ and $f(2)=3$ .

For the linear function $f(x)=mx+c$ :

$f(0)=m(0)+c=c$ . Given $f(0)=1$ , so $c=1$ .

$f'(x)=m$ (the derivative of a linear function is its constant slope). Given $f'(0)=1$ , so $m=1$ .

Hence $f(x)=x+1$ .

$f(2)=2+1=3$ .

✓Final answer
$The correct option is (a) 3.$
CBSE 2026Set ANNUAL1 markMCQ
$f'(7)$
(a) -1
(b) 1
(c) 5
(d) 7
›Reveal solutionSolution

Since $7>5$ , $f(x)=x-5$ near $x=7$ , so $f'(x)=1$ there, giving $f'(7)=1$ .

$f(x)=|x-5|=\begin{cases}x-5 & x\ge5\\ 5-x & x<5\end{cases}$

At $x=7$ (which is $>5$ ), $f(x)=x-5$ in a neighborhood of $x=7$ , so $f$ is differentiable there with $f'(x)=1$ .

$f'(7)=1$ .

✓Final answer
$The correct option is (b) 1.$
CBSE 2025Set ANNUAL1 markMCQ
$f(x) = x|x|$
(a) does not exist
(b) $6$
(c) $0$
(d) $-6$
›Reveal solutionSolution

Near $x=-3$ (which is negative), $f(x)=x|x|$ simplifies to $-x^2$ , whose derivative is $-2x$ .

Since $x=-3<0$ , in a neighbourhood of $-3$ we have $|x|=-x$ , so $f(x)=x|x|=x(-x)=-x^2$ .

Differentiating, $f'(x)=-2x$ .

At $x=-3$ : $f'(-3)=-2(-3)=6$ .

✓Final answer
$6$
CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{d}{dx}\left(\dfrac{2}{\pi}\sin x^\circ\right)$
(a) $\dfrac{\pi}{90}\cos x^\circ$
(b) $\dfrac{\pi}{180}\cos x^\circ$
(c) $\dfrac{2}{\pi}\cos x^\circ$
(d) $\dfrac{1}{90}\cos x^\circ$
›Reveal solutionSolution

Since calculus derivatives of sine require the angle in radians, first rewrite $x^\circ$ as $\dfrac{\pi x}{180}$ radians.

We have $\sin x^\circ=\sin\left(\dfrac{\pi x}{180}\right)$ . So

$\dfrac{d}{dx}\left(\dfrac{2}{\pi}\sin x^\circ\right)=\dfrac{2}{\pi}\cdot\cos\left(\dfrac{\pi x}{180}\right)\cdot\dfrac{\pi}{180}=\dfrac{2}{180}\cos x^\circ=\dfrac{1}{90}\cos x^\circ.$

✓Final answer
$\dfrac{1}{90}\cos x^\circ$
CBSE 2025Set MARCH1 markMCQ
$\frac{dy}{dx}$
(a) $nx^{n-1}$
(b) $anx^{n-1}$
(c) $0$
(d) $anx^{n+1}$
›Reveal solutionSolution

Applying the power rule, $\dfrac{d}{dx}(ax^n) = anx^{n-1}$ — option (b).

GSEB Class-12 Statistics, Differentiation chapter:

The power rule states $\dfrac{d}{dx}(x^n) = n x^{n-1}$ , and a constant multiplier is carried through:

$\frac{dy}{dx} = \frac{d}{dx}(a x^n) = a \cdot n x^{n-1} = an x^{n-1}$

✓Final answer
$anx^{n-1}$
CBSE 2023Set ANNUAL1 markMCQ
$f(x) = mx + c$
(a) 3
(b) 1
(c) 4
(d) 2
›Reveal solutionSolution

With $f(x)=mx+c$ , the condition $f(0)=1$ gives $c=1$ and $f'(0)=1$ gives $m=1$ , so $f(3)=4$ .

$f(x)=mx+c \Rightarrow f(0)=c$ . Given $f(0)=1$ , so $c=1$ .

$f'(x)=m$ for all $x$ (the derivative of a linear function is its slope, constant everywhere). Given $f'(0)=1$ , so $m=1$ .

Thus $f(x)=x+1$ , and $f(3)=3+1=4$ .

✓Final answer
$f(3)=4$
CBSE 2019Set ANNUAL1 markMCQ
$f(x)=x^2-3x$
(a) both irrational
(b) one rational and another irrational
(c) both positive integers
(d) both negative integers
›Reveal solutionSolution

Solving $x^2-3x = 2x-3$ leads to $x^2-5x+3=0$ , whose discriminant $13$ is not a perfect square, so both roots $\dfrac{5\pm\sqrt{13}}{2}$ are irrational.

$f(x)=x^2-3x$ , so $f'(x)=2x-3$ .

Set $f(x)=f'(x)$ : $x^2-3x = 2x-3$ .

Rearrange: $x^2-3x-2x+3 = 0 \Rightarrow x^2-5x+3=0$ .

Discriminant $= (-5)^2-4(1)(3) = 25-12 = 13$ , which is not a perfect square.

Roots: $x = \dfrac{5\pm\sqrt{13}}{2}$ — since $\sqrt{13}$ is irrational, both roots are irrational.

✓Final answer
$The correct option is (a) both irrational.$

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Q.If f(x)=1xnf(x)=\dfrac{1}{x^n}f(x)=xn1​, for x≠0x\ne 0x=0, n∈Nn\in Nn∈N, then prove that f′(x)=−nxn+1f'(x)=-\dfrac{n}{x^{n+1}}f′(x)=−xn+1n​.

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Q.If $f(x)=\dfrac{1}{x^n}$ , for $x\ne 0$ , $n\in N$ , then prove that $f'(x)=-\dfrac{n}{x^{n+1}}$ .

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