If f(x)=ex, then prove that f'(x)=ex.

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f(x)=e^x

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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):

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

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Let $f(x)=e^x$ , so $f(x+h)=e^{x+h}=e^x\cdot e^h$ .

$f(x+h)-f(x)=e^x(e^h-1)$ .

$\dfrac{f(x+h)-f(x)}h=e^x\cdot\dfrac{e^h-1}h$ . …

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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 $c=1$ .

…

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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 $f'(7)=1$ .

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

…

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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 $f(x)=x|x|=x(-x)=-x^2$ . …

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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 …

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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: …

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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 $c=1$ .

…

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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. …

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Q.If f(x)=exf(x)=e^xf(x)=ex, then prove that f′(x)=exf'(x)=e^xf′(x)=ex.

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Q.If $f(x)=e^x$ , then prove that $f'(x)=e^x$ .

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