If $g(x) = (x^2+2x+3)f(x)$ and $f(0) = 5$ and $\displaystyle\lim_{x\to0}\dfrac{f(x)-5}{x}=4$, then $g'(0)$ is (1) 20 (2) 14 (3) 18 (4) 12

If g(x) = (x²+2x+3)f(x) and f(0) = 5 and limx→0(f(x)-5)/(x)=4, then g'(0) is (1) 20 (2) 14 (3) 18 (4) 12

Exercise 10.5 · Q20

g(x) = (x^2+2x+3)f(x)

(1) 20

(2) 14

(3) 18

(4) 12

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

73% · 105/143 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Step 1. $g(x)=(x^2+2x+3)f(x)$ . Apply the product rule with $u=x^2+2x+3$ , $v=f(x)$ :

g'(x)=u'(x)f(x)+u(x)f'(x)=(2x+2)f(x)+(x^2+2x+3)f'(x)

Step 2. Recognize $f'(0)$ : since $f(0)=5$ , the given limit is exactly the derivative definition

f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)-5}{x}=4

Step 3. Evaluate the coefficients at $x=0$ : $u'(0)=2(0)+2=2$ and $u(0)=0+0+3=3$ .

Step 4. Substitute into the product rule:

$g'(0)=u'(0)f(0)+u(0)f'(0)=2(5)+3(4)=10+12=22$ …

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$\frac{d}{dx}(x+a) =$
›Reveal solutionSolution

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

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

…

Start your 14-day free trial to see the full solution →
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:

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

…

Start your 14-day free trial to see the full solution →
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, …

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 $\cos(x+h)-\cos x = -2\sin\left(x+\tfrac{h}{2}\right)\sin\tfrac{h}{2}$ , …

Start your 14-day free trial to see the full solution →
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 $\log e^x = x$ :

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

…

Start your 14-day free trial to see the full solution →
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 $x$ :

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

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(x^2 + 25x) =$
(a) $2x$
(b) 25
(c) $2x + 25$
(d) 0
›Reveal solutionSolution

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

…

Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →

←Previous question · 10.4.8Q19 — If f(x) = x+1, x<2 \2x-1, x≥ 2 , then f'(2) is (1) 0 (2) 1 (3) 2 (4) d…→Next question · 10.4.8Q21 — If f(x) = x+2, -1<x<3 \5, x=3 \8-x, x>3 , then at x=3, f'(x) is (1) 1 …

Q.If g(x)=(x2+2x+3)f(x)g(x) = (x^2+2x+3)f(x)g(x)=(x2+2x+3)f(x) and f(0)=5f(0) = 5f(0)=5 and lim⁡x→0f(x)−5x=4\displaystyle\lim_{x\to0}\dfrac{f(x)-5}{x}=4x→0lim​xf(x)−5​=4, then g′(0)g'(0)g′(0) is

🎓Unlock everything free for 14 days

Q.If $g(x) = (x^2+2x+3)f(x)$ and $f(0) = 5$ and $\displaystyle\lim_{x\to0}\dfrac{f(x)-5}{x}=4$ , then $g'(0)$ is

Unlock everything free for 14 days