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The concept of derivatives is explained in this exercise. In our daily life, we often observe situations where one quantity changes with respect to another, like the speed of a moving car, rise in water level in the oceans or reservoirs, or a change in prices of stocks with time. In mathematics, these concepts are explained using the concept of derivatives. Derivative is defined as the rate at which one quantity changes with respect to another or how fast or slow something is changing at a particular moment. The concept of derivatives forms the foundation for higher-level calculus.
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JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
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In NCERT, students will learn how to determine the derivatives of algebraic, polynomial, and trigonometric functions. NCERT Solutions of Exercise 12.2 are designed by subject experts in a very comprehensive and systematic manner that ensures the accuracy of the solutions. Derivatives help us to calculate how things change, which is useful in science, engineering, and economics. Check NCERT Solutions to get detailed solutions for Science and Maths from Class 6 to Class 12.
Question1: Find the derivative of $x ^ 2 -2 \: \: at \: \: x = 10$
Answer:
F(x)= $x ^ 2 -2 \: \:$
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
The derivative of f(x) at x = 10:
$f'(10)=\lim\limits_{h\rightarrow 0}\frac{f(10+h)-f(10)}{h}$
$f'(10)=\lim\limits_{h\rightarrow 0}\frac{(10+h)^2-2-((10)^2-2)}{h}$
$f'(10)=\lim\limits_{h\rightarrow 0}\frac{100+20h+h^2-2-100 +2}{h}$
$f'(10)=\lim\limits_{h\rightarrow 0}\frac{20h+h^2}{h}$
$f'(10)=\lim\limits_{h\rightarrow 0}20+h$
$f'(10)=20+0$
$f'(10)=20$
Question 2: Find the derivative of x at x = 1.
Answer:
Given
f(x)= x
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
The derivative of f(x) at x = 1:
$f'(1)=\lim\limits_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h}$
$f'(1)=\lim\limits_{h\rightarrow 0}\frac{(1+h)-(1)}{h}$
$f'(1)=\lim\limits_{h\rightarrow 0}\frac{(h)}{h}$
$f'(1)=1$ (Answer)
Question 3: Find the derivative of 99x at x = l00.
Answer:
f(x)= 99x
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
The derivative of f(x) at x = 100:
$f'(100)=\lim\limits_{h\rightarrow 0}\frac{f(100+h)-f(100)}{h}$
$f'(100)=\lim\limits_{h\rightarrow 0}\frac{99(100+h)-99(100)}{h}$
$f'(100)=\lim\limits_{h\rightarrow 0}\frac{99h}{h}$
$f'(100)=99$
Question 4: (i) Find the derivative of the following functions from first principle. $x ^3 -27$
Answer:
Given
f(x)= $x ^3 -27$
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{(x+h)^3-27-((x)^3-27)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{x^3+h^3+3x^2h+3h^2x-27+x^3+27}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{h^3+3x^2h+3h^2x}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}{h^2+3x^2+3hx}$
$f'(x)=3x^2$
Question 4:(ii) Find the derivative of the following function from first principle. $( x-1)(x-2)$
Answer:
f(x)= $( x-1)(x-2)$
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{x^2+xh-2x+hx+h^2-2h-x-h+2-x^2+2x+x-2}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{2hx+h^2-3h}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}{2x+h-3}$
$f'(x)=2x-3$ (Answer)
Question 4:(iii) Find the derivative of the following functions from first principle. $1 / x ^2$
Answer:
f(x)= $1 / x ^2$
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{1/(x+h)^2-1/(x^2)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0} \frac{\frac{x^2-(x+h)^2}{(x+h)^2x^2}}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0} \frac{x^2-x^2-2xh-h^2}{h(x+h)^2x^2}$
$f'(x)=\lim\limits_{h\rightarrow 0} \frac{-2xh-h^2}{h(x+h)^2x^2}$
$f'(x)=\lim\limits_{h\rightarrow 0} \frac{-2x-h}{(x+h)^2x^2}$
$f'(x)= \frac{-2x-0}{(x+0)^2x^2}$
$f'(x)= \frac{-2}{x^3}$ (Answer)
Question 4:(iv) Find the derivative of the following functions from first principle. $\frac{x +1}{x-1}$
Answer:
Given:
$f(x)=\frac{x +1}{x-1}$
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{x^2-x+hx-h+x-1-x^2-xh+x-x-h+1}{(x-1)(x+h-1)h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-2h}{(x-1)(x+h-1)h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{-2}{(x-1)(x+h-1)}$
$f'(x)=\frac{-2}{(x-1)(x+0-1)}$
$f'(x)=\frac{-2}{(x-1)^2}$
Answer:
$f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1$
As we know, the property,
$f'(x^n)=nx^{n-1}$
applying that property we get
$f '(x) = 100\frac{x^{99}}{100}+ 99\frac{x ^ {98}}{99} + ....+ 2\frac{x }{2}+ 1+ 0$
$f '(x) = x^{99}+x^{98}+......x+1$
Now.
$f '(0) = 0^{99}+0^{98}+......0+1=1$
$f '(1) = 1^{99}+1^{98}+......1+1=100$
So,
Here
$1\times 100=100$
$f'(0)\times 100=f'(1)$
Hence Proved.
Answer:
Given
$f(x)=x ^n + ax ^{n-1} + a ^ 2 x ^{n-2} + ....+ a ^ { n-1} x + a ^n$
As we know, the property,
$f'(x^n)=nx^{n-1}$
applying that property we get
$f'(x)=nx ^{n-1} + a(n-1)x ^{n-2} + a ^ 2(n-2) x ^{n-3} + ....+ a ^ { n-1} 1 + 0$
$f'(x)=nx ^{n-1} + a(n-1)x ^{n-2} + a ^ 2(n-2) x ^{n-3} + ....+ a ^ { n-1}$
Question 7:(i) For some constants a and b, find the derivative of $( x - a ) ( x -b )$
Answer:
Given
$f(x)=( x - a ) ( x -b )=x^2-ax-bx+ab$
As we know, the property,
$f'(x^n)=nx^{n-1}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
applying that property we get
$f'(x)=2x-a-b$
Question 7:(ii) For some constants a and b, find the derivative of $( ax ^2 + b)^2$
Answer:
Given
$f(x)=( ax ^2 + b)^2=a^2x^4+2abx^2+b^2$
As we know, the property,
$f'(x^n)=nx^{n-1}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
applying those properties we get
$f'(x)=4a^2x^3+2(2)abx+0$
$f'(x)=4a^2x^3+4abx$
$f'(x)=4ax(ax^2+b)$
Question 7:(iii) For some constants a and b, find the derivative of $\frac{x - a }{x -b }$
Answer:
Given,
$f(x)=\frac{x - a }{x -b }$
Now As we know the quotient rule of derivative,
$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$
So applying this rule, we get
$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{(x-b)\frac{d(x-a)}{dx}-(x-a)\frac{d(x-b)}{dx}}{(x-b)^2}$
$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{(x-b)-(x-a)}{(x-b)^2}$
$\frac{d(\frac{x-a}{x-b})}{dx}=\frac{a-b}{(x-b)^2}$
Hence
$f'(x)=\frac{a-b}{(x-b)^2}$
Question 8: Find the derivative of $\frac{x ^n - a ^n }{x - a }$ for some constant a.
Answer:
Given,
$f(x)=\frac{x ^n - a ^n }{x - a }$
Now As we know the quotient rule of derivative,
$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$
So applying this rule, we get
$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{(x-a)\frac{d(x^n-a^n)}{dx}-(x^n-a^n)\frac{d(x-a)}{dx}}{(x-a)^2}$
$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{(x-a)nx^{n-1}-(x^n-a^n)}{(x-a)^2}$
$\frac{d(\frac{x^n-a^n}{x-a})}{dx}=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$
Hence
$f'(x)=\frac{nx^n-anx^{n-1}-x^n+a^n}{(x-a)^2}$
Question 9:(i) Find the derivative of $2x - 3/4$
Answer:
Given:
$f(x)=2x - 3/4$
As we know, the property,
$f'(x^n)=nx^{n-1}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
applying that property we get
$f'(x)=2-0$
$f'(x)=2$
Question 9:(ii) Find the derivative of $( 5x^3 + 3x -1 ) ( x -1)$
Answer:
Given.
$f(x)=( 5x^3 + 3x -1 ) ( x -1)=5x^4+3x^2-x-5x^3-3x+1$
$f(x)=5x^4-5x^3+3x^2-4x+1$
As we know, the property,
$f'(x^n)=nx^{n-1}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
applying that property we get
$f'(x)=5(4)x^3-5(3)x^2+3(2)x-4+0$
$f'(x)=20x^3-15x^2+6x-4$
Question 9:(iii) Find the derivative of $x ^{-3} ( 5 + 3x )$
Answer:
Given
$f(x)=x ^{-3} ( 5 + 3x )=5x^{-3}+3x^{-2}$
As we know, the property,
$f'(x^n)=nx^{n-1}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
applying that property we get
$f'(x)=(-3)5x^{-4}+3(-2)x^{-3}$
$f'(x)=-15x^{-4}-6x^{-3}$
Question 9:(iv) Find the derivative of $x ^5 ( 3 - 6 x ^{-9})$
Answer:
Given
$f(x)=x ^5 ( 3 - 6 x ^{-9})=3x^5-6x^{-4}$
As we know, the property,
$f'(x^n)=nx^{n-1}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
applying that property we get
$f'(x)=(5)3x^4-6(-4)x^{-5}$
$f'(x)=15x^4+24x^{-5}$
Question 9:(v) Find the derivative of $x ^{-4} ( 3 - 4x ^{-5})$
Answer:
Given
$f(x)=x ^{-4} ( 3 - 4x ^{-5})=3x^{-4}-4x^{-9}$
As we know, the property,
$f'(x^n)=nx^{n-1}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
applying that property we get
$f'(x)=(-4)3x^{-5}-(-9)4x^{-10}$
$f'(x)=-12x^{-5}+36x^{-10}$
Question 9:(vi) Find the derivative of $\frac{2}{x+1}- \frac{x^2 }{3 x-1}$
Answer:
Given
$f(x)=\frac{2}{x+1}- \frac{x^2 }{3 x-1}$
As we know the quotient rule of derivative:
$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$
and the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
So applying this rule, we get
$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{(x+1)\frac{d(2)}{dx}-2\frac{d(x+1)}{dx}}{(x+1)^2}-\frac{(3x-1)\frac{d(x^2)}{dx}-x^2\frac{d(3x-1)}{dx}}{(3x-1)^2}$
$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{(3x-1)2x-x^23}{(3x-1)^2}$
$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{6x^2-2x-3x^2}{(3x-1)^2}$
$\frac{d(\frac{2}{x+1}-\frac{x^2}{3x-1})}{dx}=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$
Hence
$f'(x)=\frac{-2}{(x+1)^2}-\frac{3x^2-2x}{(3x-1)^2}$
Question 10: Find the derivative of $\cos x$ from first principle.
Answer:
Given,
f(x)= $\cos x$
Now, As we know, The derivative of any function at x is
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x+h)-\cos(x)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x)\cos(h)-\sin(x)\sin(h)-\cos(x)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x)\cos(h)-\cos(x)}{h}-\frac{\sin(x)\sin(h)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\frac{\cos(x)(\cos(h)-1)}{h}-\frac{\sin(x)\sin(h)}{h}$
$f'(x)=\lim\limits_{h\rightarrow 0}\cos(x)\frac{-2sin^2(h/2)}{h}\cdot -\sin(x)\frac{sinh}{h}$
$f'(x)=\cos(x)(0)-sinx(1)$
$f'(x)=-\sin(x)$
Question 11:(i) Find the derivative of the following functions: $\sin x \cos x$
Answer:
Given,
f(x)= $\sin x \cos x$
Now, As we know the product rule of derivative,
$\frac{d(y_1y_2)}{dx}=y_1\frac{dy_2}{dx}+y_2\frac{dy_1}{dx}$
So, applying the rule here,
$\frac{d(\sin x\cos x)}{dx}=\sin x\frac{d\cos x}{dx}+\cos x\frac{d\sin x}{dx}$
$\frac{d(\sin x\cos x)}{dx}=\sin x(-\sin x)+\cos x (\cos x)$
$\frac{d(\sin x\cos x)}{dx}=-\sin^2 x+\cos^2 x$
$\frac{d(\sin x\cos x)}{dx}=\cos 2x$
Question 11:(ii) Find the derivative of the following functions: $\sec x$
Answer:
Given
$f(x)=\sec x=\frac{1}{\cos x}$
Now As we know the quotient rule of derivative,
$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$
So applying this rule, we get
$\frac{d(\frac{1}{\cos x})}{dx}=\frac{\cos x\frac{d(1)}{dx}-1\frac{d(\cos x)}{dx}}{\cos ^2x}$
$\frac{d(\frac{1}{\cos x})}{dx}=\frac{-1(-\sin x)}{\cos ^2x}$
$\frac{d(\frac{1}{\cos x})}{dx}=\frac{\sin x}{\cos ^2x}=\frac{\sin x}{\cos x}\frac{1}{\cos x}$
$\frac{d(\sec x)}{dx}=\tan x\sec x$
Question 11: (iii) Find the derivative of the following functions: $5 \sec x + 4 \cos x$
Answer:
Given
$f(x)=5 \sec x + 4 \cos x$
As we know the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
Applying the property, we get
$\frac{d(5\sec x+4\cos x)}{dx}=\frac{d(5\sec x)}{dx}+\frac{d(4\cos x)}{dx}$
$\frac{d(5\sec x+4\cos x)}{dx}=5\tan x\sec x+4(-\sin x)$
$\frac{d(5\sec x+4\cos x)}{dx}=5\tan x\sec x-4\sin x$
Question 11:(iv) Find the derivative of the following functions: $\csc x$
Answer:
Given :
$f(x)=\csc x=\frac{1}{\sin x}$
Now As we know the quotient rule of derivative,
$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$
So applying this rule, we get
$\frac{d(\frac{1}{\sin x})}{dx}=\frac{(\sin x)\frac{d(1)}{dx}-1\frac{d(\sin x)}{dx}}{(\sin x)^2}$
$\frac{d(\frac{1}{\sin x})}{dx}=\frac{-1(\cos x)}{(\sin x)^2}$
$\frac{d(\frac{1}{\sin x})}{dx}=-\frac{(\cos x)}{(\sin x)}\frac{1}{\sin x}$
$\frac{d(\csc x)}{dx}=-\cot x \csc x$
Question 11:(v) Find the derivative of the following functions: $3 \cot x + 5 \csc x$
Answer:
Given,
$f(x)=3 \cot x + 5 \csc x$
As we know the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
Applying the property,
$\frac{d(3\cot x+5 \csc x)}{dx}=\frac{d(3\cot x)}{dx}+\frac{d(5\csc x)}{dx}$
$\frac{d(3\cot x+5 \csc x)}{dx}=3\frac{d(\frac{\cos x}{\sin x})}{dx}+\frac{d(5\csc x)}{dx}$
$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\frac{d(\frac{\cos x}{\sin x})}{dx}$
Now As we know the quotient rule of derivative,
$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2\frac{dy_1}{dx}-y_1\frac{dy_2}{dx}}{y_2^2}$
So applying this rule, we get
$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x\frac{d(\cos x)}{dx}-\cos x(\frac{d(\sin x)}{dx})}{\sin^2x}\right]$
$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}\right]$
$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x+3\left[\frac{-\sin^2 x-\cos^2 x}{\sin^2x}\right]$
$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\left[\frac{\sin^2 x+\cos^2 x}{\sin^2x}\right]$
$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\left[\frac{1}{\sin^2x}\right]$
$\frac{d(3\cot x+5 \csc x)}{dx}=-5\csc x\cot x-3\csc^2x$
Question 11:(vi) Find the derivative of the following functions: $5 \sin x - 6 \cos x + 7$
Answer:
Given,
$f(x)=5 \sin x - 6 \cos x + 7$
Now as we know the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
So, applying the property,
$f'(x)=5 \cos x - 6 (-\sin x ) + 0$
$f'(x)=5 \cos x + 6 (\sin x )$
$f'(x)=5 \cos x + 6 \sin x$
Question 11:(vii) Find the derivative of the following functions: $2 \tan x - 7 \sec x$
Answer:
Given
$f(x)=2 \tan x - 7 \sec x$
As we know the property
$\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$
Applying this property,
$\frac{d(2\tan x+7\sec x)}{dx}=2\frac{d\tan x}{dx}+7\frac{d\sec x}{dx}$
$\frac{d(2\tan x+7\sec x)}{dx}=2\sec^2x +7(-\sec x\tan x)$
$\frac{d(2\tan x+7\sec x)}{dx}=2\sec^2x -7\sec x\tan x$
Also read
Exercise 12.2 introduces students to the derivative of a function. In this, students will find questions that primarily revolve around the following key topics:
1) Derivatives: Derivative is the measure of the rate at which the function is changing at a given point.
Mathematically derivative of the function f(x) at point x=a is
$f^{\prime}(a)=\lim\limits _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
The instantaneous rate of change of a function at point a’ is shown by this formula.
2) Algebra of derivative of function : Rules used to find the differentiation of the combination of functions.
3) Derivative of Polynomial : $\frac{d}{d x}\left(x^n\right)=n \cdot x^{n-1}$
4) Derivatives of Trigonometric functions :
Also Read
Students can refer to the subject-wise NCERT solutions. The links to solutions are given below:
Students can access the NCERT exemplar solutions to enhance their deep understanding of the topic. These solutions are aligned with the CBSE syllabus and also help in competitive exams.
The value of f(x) at 100 =8000
The derivative of 80x is 80 which is a constant
The derivative of $80x^2=160x$ and the value at 100=16000
$Sec^2x$
The derivative of ‘a’ =0 since ‘a’ is a constant
Since the rate of change of constant =0, the derivative =0
11 questions are solved from exercise 12.2 Class 11 Maths
Definition of derivative, derivative of polynomial and trigonometric functions.
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Exam Date:01 May,2025 - 08 May,2025
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As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
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