Limits and Derivatives — Class 11 Maths Notes

Topics covered in Limits and Derivatives

1. 1.Introduction to Limits and Derivatives

   Brief Overview

   Limits and derivatives are fundamental concepts in calculus, a branch of mathematics that deals with rates of change and accumulation.

   1. Limits: A limit helps us understand the behavior of a function as it approaches a certain point. It is the value that a function approaches as the input (or variable) approaches some value.
   2. Derivatives: A derivative represents the rate at which a function is changing at any given point. It is essentially the slope of the function at that point.

   Detailed Explanation

   Limits

   A limit is written as: lim⁡𝑥→𝑎𝑓(𝑥)=𝐿limx→a​f(x)=L This means that as 𝑥x gets closer to 𝑎a, the function 𝑓(𝑥)f(x) approaches the value 𝐿L.

   Example from Real Life: Imagine you are driving a car and you start slowing down as you approach a stop sign. The limit in this case would be your speed as you get closer and closer to the stop sign. If you approach the stop sign correctly, your speed should approach zero.

   Mathematical Example: Consider the function 𝑓(𝑥)=2𝑥−1𝑥−1f(x)=x−12x−1​. To find the limit as 𝑥x approaches 1, we look at the behavior of the function near 𝑥=1x=1.

   lim⁡𝑥→12𝑥−1𝑥−1limx→1​x−12x−1​

   As 𝑥x approaches 1, the function becomes undefined because the denominator becomes zero. However, by simplifying and analyzing the function, we can find that the limit does not exist in a traditional sense, indicating a discontinuity at that point.

   Derivatives

   A derivative is written as: 𝑓′(𝑥) or 𝑑𝑦𝑑𝑥f′(x) or dxdy​ This means the rate of change of 𝑦y with respect to 𝑥x.

   Example from Real Life: Think of the speedometer in your car. It shows your speed at any given moment, which is the derivative of your position with respect to time.

   Mathematical Example: Consider the function 𝑓(𝑥)=𝑥2f(x)=x2. To find the derivative, we use the definition:

   𝑓′(𝑥)=lim⁡Δ𝑥→0𝑓(𝑥+Δ𝑥)−𝑓(𝑥)Δ𝑥f′(x)=limΔx→0​Δxf(x+Δx)−f(x)​

   For 𝑓(𝑥)=𝑥2f(x)=x2:

   𝑓′(𝑥)=lim⁡Δ𝑥→0(𝑥+Δ𝑥)2−𝑥2Δ𝑥f′(x)=limΔx→0​Δx(x+Δx)2−x2​ 𝑓′(𝑥)=lim⁡Δ𝑥→0𝑥2+2𝑥Δ𝑥+(Δ𝑥)2−𝑥2Δ𝑥f′(x)=limΔx→0​Δxx2+2xΔx+(Δx)2−x2​ 𝑓′(𝑥)=lim⁡Δ𝑥→02𝑥Δ𝑥+(Δ𝑥)2Δ𝑥f′(x)=limΔx→0​Δx2xΔx+(Δx)2​ 𝑓′(𝑥)=lim⁡Δ𝑥→0(2𝑥+Δ𝑥)f′(x)=limΔx→0​(2x+Δx) 𝑓′(𝑥)=2𝑥f′(x)=2x

   This shows that the derivative of 𝑥2x2 is 2𝑥2x, meaning the rate of change of 𝑥2x2 with respect to 𝑥x is 2𝑥2x.

   Activities and Problems

   1. Activity: Plot the function 𝑓(𝑥)=𝑥2f(x)=x2 and visually estimate the slope at various points. Compare your visual estimates with the derivative 𝑓′(𝑥)=2𝑥f′(x)=2x.

   2. Problem: Find the limit of the function 𝑔(𝑥)=𝑥2−1𝑥−1g(x)=x−1x2−1​ as 𝑥x approaches 1.

   Applications and Careers

   1. Engineering: Limits and derivatives are used in designing and analyzing systems, predicting changes, and optimizing performance.
   2. Economics: Used to calculate marginal costs and revenues, helping businesses make informed decisions.
   3. Medicine: Modeling the growth of diseases and the effect of treatments over time.

2. 2.Intuitive Idea of Derivatives

   Brief Overview

   A derivative represents the rate at which a function is changing at any given point. Think of it as a measure of how quickly something is happening. In simpler terms, it tells us the "speed" or "slope" of a function at a particular point.

   Detailed Explanation

   To understand the concept of derivatives intuitively, let's break it down with a simple real-life analogy.

   Real-Life Analogy: Speed of a Car

   Imagine you are driving a car and you look at the speedometer. The speedometer shows you how fast the car is going at that exact moment. This speed is essentially the derivative of your position with respect to time.

   • Position: This tells you where you are.
   • Speed: This tells you how fast you are moving at a particular moment.

   If you plot your position over time on a graph, the slope of the curve at any point on this graph represents your speed at that point. The steeper the slope, the faster you are going.

   Slope of a Function

   Consider a simple function 𝑦=𝑓(𝑥)y=f(x). The derivative of this function at a point 𝑥x gives us the slope of the tangent line to the function at that point.

   1. Tangent Line: A straight line that touches the curve at one point and has the same slope as the curve at that point.
   2. Slope: Measures how steep the line is.

   Mathematical Example

   Let's take the function 𝑦=𝑥2y=x2 as an example:

   1. At 𝑥=1x=1, the function 𝑦=𝑥2y=x2 becomes 𝑦=12=1y=12=1.
   2. To find the derivative (or the slope of the tangent) at 𝑥=1x=1, we use the definition of the derivative:𝑓′(𝑥)=lim⁡Δ𝑥→0𝑓(𝑥+Δ𝑥)−𝑓(𝑥)Δ𝑥f′(x)=limΔxf(x+Δx)−f(x)​For 𝑓(𝑥)=𝑥2f(x)=x2:𝑓′(𝑥)=lim⁡Δ𝑥→0(𝑥+Δ𝑥)2−𝑥2Δ𝑥f′(x)=limΔx(x+Δx)2−x2​Simplifying this, we get:𝑓′(𝑥)=lim⁡Δ𝑥→0𝑥2+2𝑥Δ𝑥+(Δ𝑥)2−𝑥2Δ𝑥f′(x)=limΔxx2+2xΔx+(Δx)2−x2​𝑓′(𝑥)=lim⁡Δ𝑥→02𝑥Δ𝑥+(Δ𝑥)2Δ𝑥f′(x)=limΔx2xΔx+(Δx)2​𝑓′(𝑥)=lim⁡Δ𝑥→0(2𝑥+Δ𝑥)f′(x)=lim(2x+Δx)𝑓′(𝑥)=2𝑥f′(x)=2xSo, the derivative of 𝑦=𝑥2y=x2 is 𝑓′(𝑥)=2𝑥f′(x)=2x. At 𝑥=1x=1, the slope is 𝑓′(1)=2×1=2f′(1)=2×1=2.

   Visualizing the Concept

   To help you visualize the concept, think of the graph of 𝑦=𝑥2y=x2. At any point on this graph, the slope of the tangent line can be found using the derivative 𝑓′(𝑥)=2𝑥f′(x)=2x.

   • When 𝑥=0x=0, the slope is 0 (the tangent line is flat).
   • When 𝑥>0x>0, the slope is positive (the tangent line goes upwards).
   • When 𝑥<0x<0, the slope is negative (the tangent line goes downwards).

   Activities and Problems

   1. Activity: Draw the graph of 𝑦=𝑥2y=x2. At different points (e.g., 𝑥=−2,−1,0,1,2x=−2,−1,0,1,2), draw the tangent lines and calculate their slopes using the derivative 𝑓′(𝑥)=2𝑥f′(x)=2x.

   2. Problem: For the function 𝑦=3𝑥3y=3x3, find the derivative and calculate the slope at 𝑥=1x=1 and 𝑥=−1x=−1.

   Applications and Careers

   1. Physics: Understanding motion, velocity, and acceleration.
   2. Economics: Analyzing cost functions and marginal rates.
   3. Engineering: Designing systems with varying rates of change, such as speed and pressure.

3. 3.Limits

   Brief Overview

   A limit helps us understand the behavior of a function as it approaches a certain point. It is the value that a function approaches as the input (or variable) approaches some value.

   Detailed Explanation

   Concept of Limits

   When we talk about limits, we are looking at what happens to a function 𝑓(𝑥)f(x) as 𝑥x gets very close to a certain value 𝑎a. The limit of 𝑓(𝑥)f(x) as 𝑥x approaches 𝑎a is denoted as:

   lim⁡𝑥→𝑎𝑓(𝑥)=𝐿limx→a​f(x)=L

   This means that as 𝑥x gets closer and closer to 𝑎a, 𝑓(𝑥)f(x) gets closer and closer to 𝐿L.

   Real-Life Analogy: Approaching a Stop Sign

   Imagine you are driving a car and you are approaching a stop sign. As you get closer to the stop sign, you start to slow down. Your speed decreases as the distance to the stop sign decreases. The limit, in this case, would be your speed as you get extremely close to the stop sign, ideally reaching zero just as you stop.

   Mathematical Examples

   Example 1: Simple Polynomial Function

   Consider the function 𝑓(𝑥)=2𝑥+3f(x)=2x+3. We want to find the limit as 𝑥x approaches 2.

   lim⁡𝑥→2(2𝑥+3)limx→2​(2x+3)

   To find this, we can simply substitute 𝑥=2x=2 into the function:

   𝑓(2)=2(2)+3=4+3=7f(2)=2(2)+3=4+3=7

   So,

   lim⁡𝑥→2(2𝑥+3)=7limx→2​(2x+3)=7

   Example 2: A Function with a Hole

   Consider the function 𝑓(𝑥)=𝑥2−1𝑥−1f(x)=x−1x2−1​. We want to find the limit as 𝑥x approaches 1.

   First, let's factorize the numerator:

   𝑓(𝑥)=(𝑥−1)(𝑥+1)𝑥−1f(x)=x−1(x−1)(x+1)​

   For 𝑥≠1x=1, we can simplify this to:

   𝑓(𝑥)=𝑥+1f(x)=x+1

   Now, we can find the limit by substituting 𝑥=1x=1:

   lim⁡𝑥→1(𝑥+1)=1+1=2limx→1​(x+1)=1+1=2

   So,

   lim⁡𝑥→1𝑥2−1𝑥−1=2limx→1​x−1x2−1​=2

   Visualizing Limits

   To visualize limits, consider the graph of a function. As you move along the x-axis towards a particular point, the y-value (or the function value) gets closer and closer to a specific number. This number is the limit.

   Activities and Problems

   1. Activity: Draw the graph of the function 𝑓(𝑥)=𝑥2−1𝑥−1f(x)=x−1x2−1​ and observe the behavior as 𝑥x approaches 1.

   2. Problem: Find the limit of 𝑔(𝑥)=𝑥2−4𝑥−2g(x)=x−2x2−4​ as 𝑥x approaches 2.

   Applications and Careers

   1. Engineering: Limits are used in analyzing systems and predicting their behavior near certain points.
   2. Economics: Used to calculate and predict trends and limits of economic functions.
   3. Computer Science: Limits are foundational in algorithms that involve continuous data and approximations.

4. 4.Algebra of Limits

   The algebra of limits involves several fundamental properties and theorems that make evaluating limits more straightforward. These rules allow us to break down complex expressions into simpler parts whose limits can be more easily determined.

   Key Theorems of Limits

   1. Limit of a Sum:

      lim⁡𝑥→𝑎[𝑓(𝑥)+𝑔(𝑥)]=lim⁡𝑥→𝑎𝑓(𝑥)+lim⁡𝑥→𝑎𝑔(𝑥)lim[f(x)+g(x)]=limf(x)+limg(x)

   2. Limit of a Difference:

      lim⁡𝑥→𝑎[𝑓(𝑥)−𝑔(𝑥)]=lim⁡𝑥→𝑎𝑓(𝑥)−lim⁡𝑥→𝑎𝑔(𝑥)lim[f(x)−g(x)]=limf(x)−limg(x)

   3. Limit of a Product:

      lim⁡𝑥→𝑎[𝑓(𝑥)⋅𝑔(𝑥)]=(lim⁡𝑥→𝑎𝑓(𝑥))⋅(lim⁡𝑥→𝑎𝑔(𝑥))lim[f(x)⋅g(x)]=(x→alim​f(x))⋅(x→alim​g(x))

   4. Limit of a Quotient:

      lim⁡𝑥→𝑎(𝑓(𝑥)𝑔(𝑥))=lim⁡𝑥→𝑎𝑓(𝑥)lim⁡𝑥→𝑎𝑔(𝑥),provided lim⁡𝑥→𝑎𝑔(𝑥)≠0lim(g(x)f(x)​)=limx→a​g(x)limx→a​f(x)​,provided limg(x)=0

   5. Limit of a Constant:

      lim⁡𝑥→𝑎𝑐=𝑐,where 𝑐 is a constantlimc=c,where c is a constant

   6. Limit of a Power:

      lim⁡𝑥→𝑎[𝑓(𝑥)]𝑛=(lim⁡𝑥→𝑎𝑓(𝑥))𝑛lim[f(x)]n=(x→alim​f(x))n

   7. Limit of a Root:

      lim⁡𝑥→𝑎𝑓(𝑥)𝑛=lim⁡𝑥→𝑎𝑓(𝑥)𝑛,provided the limit existslimnf(x)​=nx→alim​f(x)​,provided the limit exists

   Detailed Explanation with Examples

   1. Limit of a Sum

   If we have two functions 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x), the limit of their sum as 𝑥x approaches 𝑎a is the sum of their individual limits.

   Example: lim⁡𝑥→2(3𝑥+2+2𝑥2)=lim⁡𝑥→2(3𝑥+2)+lim⁡𝑥→2(2𝑥2)limx→2​(3x+2+2x2)=limx→2​(3x+2)+limx→2​(2x2)

   Calculating each limit individually: lim⁡𝑥→2(3𝑥+2)=3(2)+2=6+2=8limx→2​(3x+2)=3(2)+2=6+2=8 lim⁡𝑥→2(2𝑥2)=2(22)=2⋅4=8limx→2​(2x2)=2(22)=2⋅4=8

   So, lim⁡𝑥→2(3𝑥+2+2𝑥2)=8+8=16limx→2​(3x+2+2x2)=8+8=16

   2. Limit of a Difference

   If we have two functions 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x), the limit of their difference as 𝑥x approaches 𝑎a is the difference of their individual limits.

   Example: lim⁡𝑥→3(𝑥2−4𝑥)=lim⁡𝑥→3(𝑥2)−lim⁡𝑥→3(4𝑥)limx→3​(x2−4x)=limx→3​(x2)−limx→3​(4x)

   Calculating each limit individually: lim⁡𝑥→3(𝑥2)=32=9limx→3​(x2)=32=9 lim⁡𝑥→3(4𝑥)=4(3)=12limx→3​(4x)=4(3)=12

   So, lim⁡𝑥→3(𝑥2−4𝑥)=9−12=−3limx→3​(x2−4x)=9−12=−3

   3. Limit of a Product

   If we have two functions 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x), the limit of their product as 𝑥x approaches 𝑎a is the product of their individual limits.

   Example: lim⁡𝑥→1(𝑥⋅(𝑥+1))=lim⁡𝑥→1𝑥⋅lim⁡𝑥→1(𝑥+1)limx→1​(x⋅(x+1))=limx→1​x⋅limx→1​(x+1)

   Calculating each limit individually: lim⁡𝑥→1𝑥=1limx→1​x=1 lim⁡𝑥→1(𝑥+1)=1+1=2limx→1​(x+1)=1+1=2

   So, lim⁡𝑥→1(𝑥⋅(𝑥+1))=1⋅2=2limx→1​(x⋅(x+1))=1⋅2=2

   4. Limit of a Quotient

   If we have two functions 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x), the limit of their quotient as 𝑥x approaches 𝑎a is the quotient of their individual limits, provided the limit of the denominator is not zero.

   Example: lim⁡𝑥→2(𝑥2−4𝑥−2)limx→2​(x−2x2−4​)

   First, we simplify the function:

   𝑥2−4𝑥−2=(𝑥−2)(𝑥+2)𝑥−2=𝑥+2,𝑥≠2x−2x2−4​=x−2(x−2)(x+2)​=x+2,x=2

   So, we find the limit of the simplified function:

   lim⁡𝑥→2(𝑥+2)=2+2=4limx→2​(x+2)=2+2=4

   Thus, lim⁡𝑥→2(𝑥2−4𝑥−2)=4limx→2​(x−2x2−4​)=4

   5. Limit of a Constant

   The limit of a constant function is just the constant itself.

   Example: lim⁡𝑥→57=7limx→5​7=7

   6. Limit of a Power

   If 𝑓(𝑥)f(x) is a function and 𝑛n is a positive integer, the limit of 𝑓(𝑥)f(x) raised to the power of 𝑛n as 𝑥x approaches 𝑎a is the limit of 𝑓(𝑥)f(x) raised to the power of 𝑛n.

   Example: lim⁡𝑥→2(3𝑥)2=(lim⁡𝑥→23𝑥)2limx→2​(3x)2=(limx→2​3x)2

   Calculating the inner limit: lim⁡𝑥→23𝑥=3⋅2=6limx→2​3x=3⋅2=6

   So, (lim⁡𝑥→23𝑥)2=62=36(limx→2​3x)2=62=36

   7. Limit of a Root

   If 𝑓(𝑥)f(x) is a function and 𝑛n is a positive integer, the limit of the 𝑛nth root of 𝑓(𝑥)f(x) as 𝑥x approaches 𝑎a is the 𝑛nth root of the limit of 𝑓(𝑥)f(x), provided the limit exists.

   Example: lim⁡𝑥→9𝑥=lim⁡𝑥→9𝑥limx→9​x​=limx→9​x​

   Calculating the inner limit: lim⁡𝑥→9𝑥=9limx→9​x=9

   So, lim⁡𝑥→9𝑥=9=3limx→9​x​=9​=3

5. 5.Limits of Polynomials and Rational Functions

   Limits of Polynomials

   For any polynomial function 𝑃(𝑥)P(x), the limit as 𝑥x approaches any finite value 𝑎a is simply the value of the polynomial at 𝑎a. This is because polynomials are continuous functions.

   Theorem: Limit of a Polynomial

   If 𝑃(𝑥)P(x) is a polynomial, then: lim⁡𝑥→𝑎𝑃(𝑥)=𝑃(𝑎)limx→a​P(x)=P(a)

   Example: Consider the polynomial 𝑃(𝑥)=3𝑥3−2𝑥2+𝑥−5P(x)=3x3−2x2+x−5. To find the limit as 𝑥x approaches 2: lim⁡𝑥→2(3𝑥3−2𝑥2+𝑥−5)=3(2)3−2(2)2+2−5limx→2​(3x3−2x2+x−5)=3(2)3−2(2)2+2−5 =3(8)−2(4)+2−5=3(8)−2(4)+2−5 =24−8+2−5=24−8+2−5 =13=13

   Limits of Rational Functions

   Rational functions are ratios of polynomials. If 𝑅(𝑥)=𝑃(𝑥)𝑄(𝑥)R(x)=Q(x)P(x)​, where both 𝑃(𝑥)P(x) and 𝑄(𝑥)Q(x) are polynomials, the limit as 𝑥x approaches 𝑎a depends on whether 𝑄(𝑎)Q(a) is zero or not.

   Theorem: Limit of a Rational Function

   If 𝑅(𝑥)=𝑃(𝑥)𝑄(𝑥)R(x)=Q(x)P(x)​, then: lim⁡𝑥→𝑎𝑅(𝑥)=𝑃(𝑎)𝑄(𝑎),provided 𝑄(𝑎)≠0limx→a​R(x)=Q(a)P(a)​,provided Q(a)=0

   Example: Consider the rational function 𝑅(𝑥)=2𝑥2+3𝑥−1𝑥2−4R(x)=x2−42x2+3x−1​. To find the limit as 𝑥x approaches 3: lim⁡𝑥→32𝑥2+3𝑥−1𝑥2−4=2(3)2+3(3)−1(3)2−4limx→3​x2−42x2+3x−1​=(3)2−42(3)2+3(3)−1​ =2(9)+9−19−4=9−42(9)+9−1​ =18+9−15=518+9−1​ =265=526​

   Special Case: Indeterminate Form 0000​

   When both 𝑃(𝑎)P(a) and 𝑄(𝑎)Q(a) are zero, we encounter an indeterminate form 0000​. In such cases, we need to simplify the expression, often by factoring or using algebraic manipulation, to resolve the limit.

   Example: Consider 𝑅(𝑥)=𝑥2−1𝑥−1R(x)=x−1x2−1​. To find the limit as 𝑥x approaches 1: lim⁡𝑥→1𝑥2−1𝑥−1limx→1​x−1x2−1​ Factorize the numerator: 𝑥2−1𝑥−1=(𝑥−1)(𝑥+1)𝑥−1x−1x2−1​=x−1(x−1)(x+1)​ For 𝑥≠1x=1, this simplifies to: 𝑥+1x+1 Now, we can find the limit by substituting 𝑥=1x=1: lim⁡𝑥→1(𝑥+1)=1+1=2limx→1​(x+1)=1+1=2

   Key Theorems for Limits

   1. Sum Rule:

      lim⁡𝑥→𝑎[𝑓(𝑥)+𝑔(𝑥)]=lim⁡𝑥→𝑎𝑓(𝑥)+lim⁡𝑥→𝑎𝑔(𝑥)lim[f(x)+g(x)]=limf(x)+limg(x)

   2. Difference Rule:

      lim⁡𝑥→𝑎[𝑓(𝑥)−𝑔(𝑥)]=lim⁡𝑥→𝑎𝑓(𝑥)−lim⁡𝑥→𝑎𝑔(𝑥)lim[f(x)−g(x)]=limf(x)−limg(x)

   3. Product Rule:

      lim⁡𝑥→𝑎[𝑓(𝑥)⋅𝑔(𝑥)]=(lim⁡𝑥→𝑎𝑓(𝑥))⋅(lim⁡𝑥→𝑎𝑔(𝑥))lim[f(x)⋅g(x)]=(x→alim​f(x))⋅(x→alim​g(x))

   4. Quotient Rule:

      lim⁡𝑥→𝑎(𝑓(𝑥)𝑔(𝑥))=lim⁡𝑥→𝑎𝑓(𝑥)lim⁡𝑥→𝑎𝑔(𝑥),provided lim⁡𝑥→𝑎𝑔(𝑥)≠0lim(g(x)f(x)​)=limx→a​g(x)limx→a​f(x)​,provided limg(x)=0

   5. Power Rule:

      lim⁡𝑥→𝑎[𝑓(𝑥)]𝑛=(lim⁡𝑥→𝑎𝑓(𝑥))𝑛lim[f(x)]n=(x→alim​f(x))n

   6. Root Rule:

      lim⁡𝑥→𝑎𝑓(𝑥)𝑛=lim⁡𝑥→𝑎𝑓(𝑥)𝑛,provided the limit existslimnf(x)​=nx→alim​f(x)​,provided the limit exists

6. 6.Limits of Trigonometric Functions

   Trigonometric functions often appear in calculus problems, and understanding their limits is crucial. Here are some key theorems and rules related to the limits of trigonometric functions.

   Key Theorems

   1. Limit of sin⁡(𝑥)sin(x) as 𝑥x approaches 0:

      lim⁡𝑥→0sin⁡(𝑥)=0limsin(x)=0

   2. Limit of cos⁡(𝑥)cos(x) as 𝑥x approaches 0:

      lim⁡𝑥→0cos⁡(𝑥)=1limcos(x)=1

   3. Limit of tan⁡(𝑥)tan(x) as 𝑥x approaches 0:

      lim⁡𝑥→0tan⁡(𝑥)=0limtan(x)=0

   4. Limit of sin⁡(𝑥)𝑥xsin(x)​ as 𝑥x approaches 0:

      lim⁡𝑥→0sin⁡(𝑥)𝑥=1limxsin(x)​=1

   5. Limit of 1−cos⁡(𝑥)𝑥x1−cos(x)​ as 𝑥x approaches 0:

      lim⁡𝑥→01−cos⁡(𝑥)𝑥=0limx1−cos(x)​=0

   Detailed Explanation with Examples

   1. Limit of sin⁡(𝑥)sin(x) as 𝑥x approaches 0

   The sine function sin⁡(𝑥)sin(x) is continuous everywhere, including at 0. Thus: lim⁡𝑥→0sin⁡(𝑥)=sin⁡(0)=0limx→0​sin(x)=sin(0)=0

   2. Limit of cos⁡(𝑥)cos(x) as 𝑥x approaches 0

   The cosine function cos⁡(𝑥)cos(x) is also continuous everywhere, including at 0. Thus: lim⁡𝑥→0cos⁡(𝑥)=cos⁡(0)=1limx→0​cos(x)=cos(0)=1

   3. Limit of tan⁡(𝑥)tan(x) as 𝑥x approaches 0

   The tangent function tan⁡(𝑥)=sin⁡(𝑥)cos⁡(𝑥)tan(x)=cos(x)sin(x)​. Since sin⁡(𝑥)sin(x) and cos⁡(𝑥)cos(x) are both continuous at 0, and cos⁡(0)=1≠0cos(0)=1=0, we have: lim⁡𝑥→0tan⁡(𝑥)=lim⁡𝑥→0sin⁡(𝑥)lim⁡𝑥→0cos⁡(𝑥)=01=0limx→0​tan(x)=limx→0​cos(x)limx→0​sin(x)​=10​=0

   4. Limit of sin⁡(𝑥)𝑥xsin(x)​ as 𝑥x approaches 0

   This is a fundamental limit in calculus, often proven using the Squeeze Theorem or L'Hôpital's Rule. The result is: lim⁡𝑥→0sin⁡(𝑥)𝑥=1limx→0​xsin(x)​=1

   Proof Using the Squeeze Theorem: Consider the inequality: cos⁡(𝑥)≤sin⁡(𝑥)𝑥≤1 for 𝑥>0cos(x)≤xsin(x)​≤1 for x>0 Taking the limit as 𝑥x approaches 0, and knowing that cos⁡(0)=1cos(0)=1, we get: 1≤lim⁡𝑥→0sin⁡(𝑥)𝑥≤11≤limx→0​xsin(x)​≤1 Thus, lim⁡𝑥→0sin⁡(𝑥)𝑥=1limx→0​xsin(x)​=1

   5. Limit of 1−cos⁡(𝑥)𝑥x1−cos(x)​ as 𝑥x approaches 0

   This limit can be found using a trigonometric identity and L'Hôpital's Rule. The result is: lim⁡𝑥→01−cos⁡(𝑥)𝑥=0limx→0​x1−cos(x)​=0

   Proof Using L'Hôpital's Rule: Rewrite the limit in a form suitable for L'Hôpital's Rule: lim⁡𝑥→01−cos⁡(𝑥)𝑥=lim⁡𝑥→000limx→0​x1−cos(x)​=limx→0​00​ Applying L'Hôpital's Rule: lim⁡𝑥→01−cos⁡(𝑥)𝑥=lim⁡𝑥→0sin⁡(𝑥)1=sin⁡(0)=0limx→0​x1−cos(x)​=limx→0​1sin(x)​=sin(0)=0

7. 7.Exercise Question

   1. lim⁡𝑥→3(𝑥+3)limx→3​(x+3)

   This is a simple polynomial function. lim⁡𝑥→3(𝑥+3)=3+3=6limx→3​(x+3)=3+3=6

   2. lim⁡𝑥→1(𝑥−227)limx→1​(x−722​)

   This is another simple polynomial function. lim⁡𝑥→1(𝑥−227)=1−227=1−3.142857=−2.142857limx→1​(x−722​)=1−722​=1−3.142857=−2.142857

   3. lim⁡𝑥→1𝜋𝑥2limx→1​πx2

   lim⁡𝑥→1𝜋𝑥2=𝜋(1)2=𝜋limx→1​πx2=π(1)2=π

   4. lim⁡𝑥→44𝑥+3𝑥−2limx→4​x−24x+3​

   lim⁡𝑥→44𝑥+3𝑥−2=4(4)+34−2=16+32=192=9.5limx→4​x−24x+3​=4−24(4)+3​=216+3​=219​=9.5

   5. lim⁡𝑥→1𝑥10+𝑥5+1𝑥limx→1​xx10+x5+1​

   lim⁡𝑥→1𝑥10+𝑥5+1𝑥=(1)10+(1)5+11=1+1+11=3limx→1​xx10+x5+1​=1(1)10+(1)5+1​=11+1+1​=3

   6. lim⁡𝑥→0(𝑥+1)5−1𝑥limx→0​x(x+1)5−1​

   This is in the indeterminate form 0000​. We can use L'Hôpital's Rule: lim⁡𝑥→0(𝑥+1)5−1𝑥=lim⁡𝑥→05(𝑥+1)41=5(1)4=5limx→0​x(x+1)5−1​=limx→0​15(x+1)4​=5(1)4=5

   7. lim⁡𝑥→23𝑥2−𝑥−10𝑥2−4limx→2​x2−43x2−x−10​

   Factorize the numerator and the denominator: 3𝑥2−𝑥−10=(3𝑥+5)(𝑥−2)3x2−x−10=(3x+5)(x−2) 𝑥2−4=(𝑥−2)(𝑥+2)x2−4=(x−2)(x+2) lim⁡𝑥→2(3𝑥+5)(𝑥−2)(𝑥−2)(𝑥+2)=lim⁡𝑥→23𝑥+5𝑥+2=3(2)+52+2=6+54=114=2.75limx→2​(x−2)(x+2)(3x+5)(x−2)​=limx→2​x+23x+5​=2+23(2)+5​=46+5​=411​=2.75

   8. lim⁡𝑥→3𝑥4−812𝑥2−5𝑥−3limx→3​2x2−5x−3x4−81​

   Factorize the numerator and the denominator: 𝑥4−81=(𝑥2−9)(𝑥2+9)=(𝑥−3)(𝑥+3)(𝑥2+9)x4−81=(x2−9)(x2+9)=(x−3)(x+3)(x2+9) 2𝑥2−5𝑥−3=(2𝑥+1)(𝑥−3)2x2−5x−3=(2x+1)(x−3) lim⁡𝑥→3(𝑥−3)(𝑥+3)(𝑥2+9)(2𝑥+1)(𝑥−3)=lim⁡𝑥→3(𝑥+3)(𝑥2+9)2𝑥+1=(3+3)(32+9)2(3)+1=6(9+9)6+1=6⋅187=1087≈15.428571limx→3​(2x+1)(x−3)(x−3)(x+3)(x2+9)​=limx→3​2x+1(x+3)(x2+9)​=2(3)+1(3+3)(32+9)​=6+16(9+9)​=76⋅18​=7108​≈15.428571

   9. lim⁡𝑥→0𝑎𝑥+𝑏𝑐𝑥+1limx→0​cx+1ax+b​

   Substitute 𝑥=0x=0: lim⁡𝑥→0𝑎𝑥+𝑏𝑐𝑥+1=𝑎(0)+𝑏𝑐(0)+1=𝑏1=𝑏limx→0​cx+1ax+b​=c(0)+1a(0)+b​=1b​=b

   Question 10

   Given:

   lim⁡𝑥→1𝑓(𝑥)−2𝑥2−1=𝜋limx2−1f(x)−2​=π

   We need to evaluate lim⁡𝑥→1𝑓(𝑥)limx→1​f(x).

   Let's use the given limit:

   lim⁡𝑥→1𝑓(𝑥)−2𝑥2−1=𝜋limx2−1f(x)−2​=π

   We know that 𝑥2−1x2−1 can be factored as:

   𝑥2−1=(𝑥−1)(𝑥+1)x2−1=(x−1)(x+1)

   So, the given limit becomes:

   lim⁡𝑥→1𝑓(𝑥)−2(𝑥−1)(𝑥+1)=𝜋lim(x−1)(x+1)f(x)−2​=π

   As 𝑥x approaches 1, we can simplify this by using L'Hôpital's Rule. The numerator 𝑓(𝑥)−2f(x)−2 must approach 0 as 𝑥x approaches 1 for the limit to be finite. Therefore:

   𝑓(1)−2=0 ⟹ 𝑓(1)=2f(1)−2=0⟹f(1)=2

   So, lim⁡𝑥→1𝑓(𝑥)=2limx→1​f(x)=2.

   Question 11

   Given:

   𝑓(𝑥)={𝑚𝑥2+𝑛,𝑥<0𝑛𝑥+𝑚,0≤𝑥≤1𝑛𝑥3+𝑚,𝑥>1f(x)=⎩⎨⎧​mx2+n,nx+m,nx3+m,​x<00≤x≤1x>1​

   We need to find the integers 𝑚m and 𝑛n such that both lim⁡𝑥→0𝑓(𝑥)limx→0​f(x) and lim⁡𝑥→1𝑓(𝑥)limx→1​f(x) exist.

   For lim⁡𝑥→0𝑓(𝑥)limx→0​f(x) to exist:

   The left-hand limit as 𝑥→0−x→0− should equal the right-hand limit as 𝑥→0+x→0+:

   lim⁡𝑥→0−𝑓(𝑥)=lim⁡𝑥→0+𝑓(𝑥)limf(x)=limf(x)

   For 𝑥<0x<0, 𝑓(𝑥)=𝑚𝑥2+𝑛f(x)=mx2+n. So:

   lim⁡𝑥→0−𝑓(𝑥)=𝑚(0)2+𝑛=𝑛limf(x)=m(0)2+n=n

   For 0≤𝑥≤10≤x≤1, 𝑓(𝑥)=𝑛𝑥+𝑚f(x)=nx+m. So:

   lim⁡𝑥→0+𝑓(𝑥)=𝑛(0)+𝑚=𝑚limf(x)=n(0)+m=m

   For the limit to exist at 𝑥=0x=0:

   𝑛=𝑚n=m

   For lim⁡𝑥→1𝑓(𝑥)limx→1​f(x) to exist:

   The left-hand limit as 𝑥→1−x→1− should equal the right-hand limit as 𝑥→1+x→1+:

   lim⁡𝑥→1−𝑓(𝑥)=lim⁡𝑥→1+𝑓(𝑥)limf(x)=limf(x)

   For 0≤𝑥≤10≤x≤1, 𝑓(𝑥)=𝑛𝑥+𝑚f(x)=nx+m. So:

   lim⁡𝑥→1−𝑓(𝑥)=𝑛(1)+𝑚=𝑛+𝑚limf(x)=n(1)+m=n+m

   For 𝑥>1x>1, 𝑓(𝑥)=𝑛𝑥3+𝑚f(x)=nx3+m. So:

   lim⁡𝑥→1+𝑓(𝑥)=𝑛(1)3+𝑚=𝑛+𝑚limf(x)=n(1)3+m=n+m

   For the limit to exist at 𝑥=1x=1:

   𝑛+𝑚=𝑛+𝑚n+m=n+m

   Therefore, the condition 𝑛=𝑚n=m obtained earlier is sufficient for both limits to exist.

8. 8.Derivatives: Definitions and Key Concepts

   1. Definition of a Derivative

   The derivative of a function measures how the function's output value changes as its input value changes. It represents the rate of change or the slope of the function at any given point.

   Mathematically, the derivative of a function 𝑓(𝑥)f(x) at a point 𝑥=𝑎x=a is defined as: 𝑓′(𝑎)=lim⁡ℎ→0𝑓(𝑎+ℎ)−𝑓(𝑎)ℎf′(a)=limh→0​hf(a+h)−f(a)​

   2. Geometric Interpretation

   The derivative at a point is the slope of the tangent line to the function at that point. If you plot the function 𝑓(𝑥)f(x) on a graph, the derivative at 𝑥=𝑎x=a gives you the slope of the line that just touches (is tangent to) the curve at that point.

   3. Notation

   There are several notations for derivatives, including:

   • 𝑓′(𝑥)f′(x)
   • 𝑑𝑓𝑑𝑥dxdf​
   • 𝐷𝑓(𝑥)Df(x)
   • 𝑦′y′ if 𝑦=𝑓(𝑥)y=f(x)

   4. Basic Rules of Differentiation

   Here are some fundamental rules for finding derivatives:

   a. Constant Rule

   The derivative of a constant function is zero. 𝑑𝑑𝑥(𝑐)=0dxd​(c)=0

   b. Power Rule

   For any real number 𝑛n: 𝑑𝑑𝑥(𝑥𝑛)=𝑛𝑥𝑛−1dxd​(xn)=nxn−1

   c. Sum/Difference Rule

   The derivative of a sum/difference is the sum/difference of the derivatives. 𝑑𝑑𝑥[𝑓(𝑥)±𝑔(𝑥)]=𝑓′(𝑥)±𝑔′(𝑥)dxd​[f(x)±g(x)]=f′(x)±g′(x)

   d. Product Rule

   The derivative of a product of two functions is given by: 𝑑𝑑𝑥[𝑓(𝑥)𝑔(𝑥)]=𝑓′(𝑥)𝑔(𝑥)+𝑓(𝑥)𝑔′(𝑥)dxd​[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)

   e. Quotient Rule

   The derivative of a quotient of two functions is given by: 𝑑𝑑𝑥[𝑓(𝑥)𝑔(𝑥)]=𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)[𝑔(𝑥)]2dxd​[g(x)f(x)​]=[g(x)]2f′(x)g(x)−f(x)g′(x)​

   f. Chain Rule

   The derivative of a composite function is given by: 𝑑𝑑𝑥[𝑓(𝑔(𝑥))]=𝑓′(𝑔(𝑥))𝑔′(𝑥)dxd​[f(g(x))]=f′(g(x))g′(x)

   5. Higher-Order Derivatives

   The second derivative, denoted as 𝑓′′(𝑥)f′′(x) or 𝑑2𝑓𝑑𝑥2dx2d2f​, measures the rate of change of the rate of change, i.e., the curvature of the function. Similarly, higher-order derivatives can be computed iteratively.

   Examples

   1. Derivative of a Polynomial: For 𝑓(𝑥)=3𝑥2+5𝑥−7f(x)=3x2+5x−7: 𝑓′(𝑥)=6𝑥+5f′(x)=6x+5

   2. Using the Product Rule: For 𝑓(𝑥)=𝑥2sin⁡(𝑥)f(x)=x2sin(x): 𝑓′(𝑥)=2𝑥sin⁡(𝑥)+𝑥2cos⁡(𝑥)f′(x)=2xsin(x)+x2cos(x)

   3. Using the Chain Rule: For 𝑓(𝑥)=(3𝑥2+2)4f(x)=(3x2+2)4: Let 𝑢=3𝑥2+2u=3x2+2, then 𝑓(𝑥)=𝑢4f(x)=u4. 𝑑𝑑𝑥(𝑢4)=4𝑢3⋅𝑑𝑢𝑑𝑥=4(3𝑥2+2)3⋅6𝑥=24𝑥(3𝑥2+2)3dxd​(u4)=4u3⋅dxdu​=4(3x2+2)3⋅6x=24x(3x2+2)3

   Applications of Derivatives

   1. Finding Tangents and Normals: Derivatives are used to find the equations of tangent and normal lines to curves.
   2. Optimization: Derivatives are used in finding local maxima and minima of functions, crucial in optimization problems.
   3. Motion: In physics, derivatives describe the velocity and acceleration of moving objects.
   4. Economics: Derivatives help in finding marginal cost and marginal revenue functions.

9. 9.Algebra of derivative of functions - theorem

   The algebra of derivatives involves rules for finding the derivatives of algebraic combinations of functions. These rules make it easier to compute derivatives of more complex functions by using the derivatives of simpler functions. The main rules include the sum rule, difference rule, product rule, and quotient rule.

   1. Sum Rule

   Statement: If 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x) are both differentiable functions, then the derivative of their sum is the sum of their derivatives.

   Formula:

   𝑑𝑑𝑥[𝑓(𝑥)+𝑔(𝑥)]=𝑓′(𝑥)+𝑔′(𝑥)dxd​[f(x)+g(x)]=f′(x)+g′(x)

   Example: If 𝑓(𝑥)=𝑥2f(x)=x2 and 𝑔(𝑥)=sin⁡(𝑥)g(x)=sin(x), then:

   𝑑𝑑𝑥[𝑥2+sin⁡(𝑥)]=𝑑𝑑𝑥[𝑥2]+𝑑𝑑𝑥[sin⁡(𝑥)]=2𝑥+cos⁡(𝑥)dxd​[x2+sin(x)]=dxd​[x2]+dxd​[sin(x)]=2x+cos(x)

   2. Difference Rule

   Statement: If 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x) are both differentiable functions, then the derivative of their difference is the difference of their derivatives.

   Formula:

   𝑑𝑑𝑥[𝑓(𝑥)−𝑔(𝑥)]=𝑓′(𝑥)−𝑔′(𝑥)dxd​[f(x)−g(x)]=f′(x)−g′(x)

   Example: If 𝑓(𝑥)=𝑥2f(x)=x2 and 𝑔(𝑥)=sin⁡(𝑥)g(x)=sin(x), then:

   𝑑𝑑𝑥[𝑥2−sin⁡(𝑥)]=𝑑𝑑𝑥[𝑥2]−𝑑𝑑𝑥[sin⁡(𝑥)]=2𝑥−cos⁡(𝑥)dxd​[x2−sin(x)]=dxd​[x2]−dxd​[sin(x)]=2x−cos(x)

   3. Product Rule

   Statement: If 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x) are both differentiable functions, then the derivative of their product is given by:

   Formula:

   𝑑𝑑𝑥[𝑓(𝑥)⋅𝑔(𝑥)]=𝑓′(𝑥)𝑔(𝑥)+𝑓(𝑥)𝑔′(𝑥)dxd​[f(x)⋅g(x)]=f′(x)g(x)+f(x)g′(x)

   Example: If 𝑓(𝑥)=𝑥2f(x)=x2 and 𝑔(𝑥)=sin⁡(𝑥)g(x)=sin(x), then:

   𝑑𝑑𝑥[𝑥2sin⁡(𝑥)]=𝑑𝑑𝑥[𝑥2]⋅sin⁡(𝑥)+𝑥2⋅𝑑𝑑𝑥[sin⁡(𝑥)]=2𝑥sin⁡(𝑥)+𝑥2cos⁡(𝑥)dxd​[x2sin(x)]=dxd​[x2]⋅sin(x)+x2⋅dxd​[sin(x)]=2xsin(x)+x2cos(x)

   4. Quotient Rule

   Statement: If 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x) are both differentiable functions and 𝑔(𝑥)≠0g(x)=0, then the derivative of their quotient is given by:

   Formula:

   𝑑𝑑𝑥[𝑓(𝑥)𝑔(𝑥)]=𝑓′(𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)[𝑔(𝑥)]2dxd​[g(x)f(x)​]=[g(x)]2f′(x)g(x)−f(x)g′(x)​

   Example: If 𝑓(𝑥)=𝑥2f(x)=x2 and 𝑔(𝑥)=sin⁡(𝑥)g(x)=sin(x), then:

   𝑑𝑑𝑥[𝑥2sin⁡(𝑥)]=2𝑥sin⁡(𝑥)−𝑥2cos⁡(𝑥)[sin⁡(𝑥)]2dxd​[sin(x)x2​]=[sin(x)]22xsin(x)−x2cos(x)​

   5. Chain Rule

   Statement: If 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x) are both differentiable functions, and if 𝑦=𝑓(𝑔(𝑥))y=f(g(x)), then the derivative of 𝑦y with respect to 𝑥x is the product of the derivative of 𝑓f with respect to 𝑔g and the derivative of 𝑔g with respect to 𝑥x.

   Formula:

   𝑑𝑦𝑑𝑥=𝑑𝑓𝑑𝑔⋅𝑑𝑔𝑑𝑥dxdy​=dgdf​⋅dxdg​

   Example: If 𝑓(𝑥)=sin⁡(𝑥)f(x)=sin(x) and 𝑔(𝑥)=𝑥2g(x)=x2, then:

   𝑑𝑑𝑥[sin⁡(𝑥2)]=cos⁡(𝑥2)⋅𝑑𝑑𝑥[𝑥2]=cos⁡(𝑥2)⋅2𝑥=2𝑥cos⁡(𝑥2)dxd​[sin(x2)]=cos(x2)⋅dxd​[x2]=cos(x2)⋅2x=2xcos(x2)

10. 10.Derivative of Polynomials and Trigonometric Functions - Theorems

    1. Derivative of Polynomials

    The process of finding the derivative of a polynomial function involves applying the power rule to each term of the polynomial.

    Theorem: Power Rule for Derivatives

    If 𝑓(𝑥)=𝑥𝑛f(x)=xn, where 𝑛n is a real number, then: 𝑑𝑑𝑥(𝑥𝑛)=𝑛𝑥𝑛−1dxd​(xn)=nxn−1

    Derivative of a General Polynomial

    For a general polynomial function 𝑃(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+…+𝑎1𝑥+𝑎0P(x)=an​xn+an−1​xn−1+…+a1​x+a0​, the derivative is: 𝑑𝑑𝑥𝑃(𝑥)=𝑎𝑛𝑑𝑑𝑥(𝑥𝑛)+𝑎𝑛−1𝑑𝑑𝑥(𝑥𝑛−1)+…+𝑎1𝑑𝑑𝑥(𝑥)+𝑑𝑑𝑥(𝑎0)dxd​P(x)=an​dxd​(xn)+an−1​dxd​(xn−1)+…+a1​dxd​(x)+dxd​(a0​) Using the power rule: 𝑃′(𝑥)=𝑎𝑛𝑛𝑥𝑛−1+𝑎𝑛−1(𝑛−1)𝑥𝑛−2+…+𝑎1⋅1P′(x)=an​nxn−1+an−1​(n−1)xn−2+…+a1​⋅1

    Example: For 𝑓(𝑥)=3𝑥4−5𝑥3+2𝑥2−4𝑥+7f(x)=3x4−5x3+2x2−4x+7: 𝑓′(𝑥)=3⋅4𝑥3−5⋅3𝑥2+2⋅2𝑥−4⋅1=12𝑥3−15𝑥2+4𝑥−4f′(x)=3⋅4x3−5⋅3x2+2⋅2x−4⋅1=12x3−15x2+4x−4

    2. Derivative of Trigonometric Functions

    Theorem: Derivatives of Basic Trigonometric Functions

    1. 𝑑𝑑𝑥(sin⁡𝑥)=cos⁡𝑥dxd​(sinx)=cosx
    2. 𝑑𝑑𝑥(cos⁡𝑥)=−sin⁡𝑥dxd​(cosx)=−sinx
    3. 𝑑𝑑𝑥(tan⁡𝑥)=sec⁡2𝑥dxd​(tanx)=sec2x
    4. 𝑑𝑑𝑥(cot⁡𝑥)=−csc⁡2𝑥dxd​(cotx)=−csc2x
    5. 𝑑𝑑𝑥(sec⁡𝑥)=sec⁡𝑥tan⁡𝑥dxd​(secx)=secxtanx
    6. 𝑑𝑑𝑥(csc⁡𝑥)=−csc⁡𝑥cot⁡𝑥dxd​(cscx)=−cscxcotx

    Proofs of Basic Trigonometric Derivatives:

    1. Derivative of sin⁡𝑥sinx:

       𝑑𝑑𝑥(sin⁡𝑥)=lim⁡ℎ→0sin⁡(𝑥+ℎ)−sin⁡𝑥ℎdxd​(sinx)=limhsin(x+h)−sinx​

       Using the sum-to-product identities:

       sin⁡(𝑥+ℎ)=sin⁡𝑥cos⁡ℎ+cos⁡𝑥sin⁡ℎsin(x+h)=sinxcosh+cosxsinh

       Substituting:

       sin⁡(𝑥+ℎ)−sin⁡𝑥ℎ=sin⁡𝑥cos⁡ℎ+cos⁡𝑥sin⁡ℎ−sin⁡𝑥ℎhsin(x+h)−sinx​=hsinxcosh+cosxsinh−sinx​=sin⁡𝑥(cos⁡ℎ−1ℎ)+cos⁡𝑥(sin⁡ℎℎ)=sinx(hcosh−1​)+cosx(hsinh​)

       As ℎ→0h→0, cos⁡ℎ−1ℎ→0hcosh−1​→0 and sin⁡ℎℎ→1hsinh​→1, so:

       𝑑𝑑𝑥(sin⁡𝑥)=cos⁡𝑥dxd​(sinx)=cosx

    2. Derivative of cos⁡𝑥cosx:

       𝑑𝑑𝑥(cos⁡𝑥)=lim⁡ℎ→0cos⁡(𝑥+ℎ)−cos⁡𝑥ℎdxd​(cosx)=limhcos(x+h)−cosx​

       Using the sum-to-product identities:

       cos⁡(𝑥+ℎ)=cos⁡𝑥cos⁡ℎ−sin⁡𝑥sin⁡ℎcos(x+h)=cosxcosh−sinxsinh

       Substituting:

       cos⁡(𝑥+ℎ)−cos⁡𝑥ℎ=cos⁡𝑥cos⁡ℎ−sin⁡𝑥sin⁡ℎ−cos⁡𝑥ℎhcos(x+h)−cosx​=hcosxcosh−sinxsinh−cosx​=cos⁡𝑥(cos⁡ℎ−1ℎ)−sin⁡𝑥(sin⁡ℎℎ)=cosx(hcosh−1​)−sinx(hsinh​)

       As ℎ→0h→0, cos⁡ℎ−1ℎ→0hcosh−1​→0 and sin⁡ℎℎ→1hsinh​→1, so:

       𝑑𝑑𝑥(cos⁡𝑥)=−sin⁡𝑥dxd​(cosx)=−sinx

    Examples

    1. Derivative of sin⁡(3𝑥)sin(3x): Using the chain rule:

       𝑑𝑑𝑥(sin⁡(3𝑥))=cos⁡(3𝑥)⋅𝑑𝑑𝑥(3𝑥)=3cos⁡(3𝑥)dxd​(sin(3x))=cos(3x)⋅dxd​(3x)=3cos(3x)

    2. Derivative of 𝑥2sin⁡𝑥x2sinx: Using the product rule:

       𝑑𝑑𝑥(𝑥2sin⁡𝑥)=𝑥2𝑑𝑑𝑥(sin⁡𝑥)+sin⁡𝑥𝑑𝑑𝑥(𝑥2)=𝑥2cos⁡𝑥+2𝑥sin⁡𝑥dxd​(x2sinx)=x2dxd​(sinx)+sinxdxd​(x2)=x2cosx+2xsinx

11. 11.Exercise Question

    1. Find the derivative of

    𝑥2−2x2−2 at 𝑥=10x=10.

    The function given is 𝑓(𝑥)=𝑥2−2f(x)=x2−2.

    To find the derivative 𝑓′(𝑥)f′(x), we use the power rule: 𝑑𝑑𝑥𝑥𝑛=𝑛𝑥𝑛−1dxd​xn=nxn−1.

    𝑓′(𝑥)=𝑑𝑑𝑥(𝑥2)−𝑑𝑑𝑥(2)=2𝑥−0=2𝑥f′(x)=dxd​(x2)−dxd​(2)=2x−0=2x

    Now, evaluate this derivative at 𝑥=10x=10:

    𝑓′(10)=2(10)=20f′(10)=2(10)=20

    So, the derivative of 𝑥2−2x2−2 at 𝑥=10x=10 is 2020.

    2. Find the derivative of 𝑥x at 𝑥=1x=1.

    The function given is 𝑓(𝑥)=𝑥f(x)=x.

    The derivative of 𝑥x is simply 11, since:

    𝑑𝑑𝑥𝑥=1dxd​x=1

    So, the derivative of 𝑥x at 𝑥=1x=1 is 11.

    3. Find the derivative of 99𝑥99x at 𝑥=100x=100.

    The function given is 𝑓(𝑥)=99𝑥f(x)=99x.

    The derivative of 99𝑥99x is 9999, since:

    𝑑𝑑𝑥(99𝑥)=99dxd​(99x)=99

    So, the derivative of 99𝑥99x at 𝑥=100x=100 is 9999.

    4. Find the derivative of the following functions from first principles.

    (i) 𝑥3−27x3−27

    Using the first principle (definition of the derivative):

    𝑓′(𝑥)=lim⁡ℎ→0𝑓(𝑥+ℎ)−𝑓(𝑥)ℎf′(x)=limh→0​hf(x+h)−f(x)​

    For 𝑓(𝑥)=𝑥3−27f(x)=x3−27:

    𝑓′(𝑥)=lim⁡ℎ→0(𝑥+ℎ)3−27−(𝑥3−27)ℎ=lim⁡ℎ→0𝑥3+3𝑥2ℎ+3𝑥ℎ2+ℎ3−𝑥3ℎ=lim⁡ℎ→03𝑥2ℎ+3𝑥ℎ2+ℎ3ℎ=lim⁡ℎ→0(3𝑥2+3𝑥ℎ+ℎ2)=3𝑥2f′(x)=limh→0​h(x+h)3−27−(x3−27)​=limh→0​hx3+3x2h+3xh2+h3−x3​=limh→0​h3x2h+3xh2+h3​=limh→0​(3x2+3xh+h2)=3x2

    So, the derivative of 𝑥3−27x3−27 is 3𝑥23x2.

    (ii) (𝑥−1)(𝑥−2)(x−1)(x−2)

    First, expand the function:

    𝑓(𝑥)=(𝑥−1)(𝑥−2)=𝑥2−3𝑥+2f(x)=(x−1)(x−2)=x2−3x+2

    Now, using the power rule:

    𝑓′(𝑥)=𝑑𝑑𝑥(𝑥2−3𝑥+2)=2𝑥−3f′(x)=dxd​(x2−3x+2)=2x−3

    So, the derivative of (𝑥−1)(𝑥−2)(x−1)(x−2) is 2𝑥−32x−3.

    (iii) 1𝑥2x21​

    Rewrite the function as 𝑓(𝑥)=𝑥−2f(x)=x−2.

    Using the power rule:

    𝑓′(𝑥)=𝑑𝑑𝑥(𝑥−2)=−2𝑥−3=−2𝑥3f′(x)=dxd​(x−2)=−2x−3=−x32​

    So, the derivative of 1𝑥2x21​ is −2𝑥3−x32​.

    (iv) 𝑥+1𝑥−1x−1x+1​

    Using the quotient rule (𝑢𝑣)′=𝑢′𝑣−𝑢𝑣′𝑣2(vu​)′=v2u′v−uv′​, where 𝑢=𝑥+1u=x+1 and 𝑣=𝑥−1v=x−1:

    𝑢=𝑥+1⇒𝑢′=1u=x+1⇒u′=1 𝑣=𝑥−1⇒𝑣′=1v=x−1⇒v′=1

    𝑓′(𝑥)=(𝑥+1)′(𝑥−1)−(𝑥+1)(𝑥−1)′(𝑥−1)2=1(𝑥−1)−(𝑥+1)1(𝑥−1)2=𝑥−1−𝑥−1(𝑥−1)2=−2(𝑥−1)2f′(x)=(x−1)2(x+1)′(x−1)−(x+1)(x−1)′​=(x−1)21(x−1)−(x+1)1​=(x−1)2x−1−x−1​=(x−1)2−2​

    So, the derivative of 𝑥+1𝑥−1x−1x+1​ is −2(𝑥−1)2−(x−1)22​.

    Question 5

    Find the derivative of 𝑥𝑛+𝑎𝑥𝑛−1+𝑎2𝑥𝑛−2+…+𝑎𝑛−1𝑥+𝑎𝑛xn+axn−1+a2xn−2+…+an−1x+an for some fixed real number 𝑎a.

    To differentiate the polynomial, we apply the power rule to each term.

    Solution:

    𝑑𝑑𝑥(𝑥𝑛+𝑎𝑥𝑛−1+𝑎2𝑥𝑛−2+…+𝑎𝑛−1𝑥+𝑎𝑛)dxd​(xn+axn−1+a2xn−2+…+an−1x+an)

    Differentiating term-by-term:

    =𝑑𝑑𝑥(𝑥𝑛)+𝑑𝑑𝑥(𝑎𝑥𝑛−1)+𝑑𝑑𝑥(𝑎2𝑥𝑛−2)+…+𝑑𝑑𝑥(𝑎𝑛−1𝑥)+𝑑𝑑𝑥(𝑎𝑛)=dxd​(xn)+dxd​(axn−1)+dxd​(a2xn−2)+…+dxd​(an−1x)+dxd​(an)=𝑛𝑥𝑛−1+𝑎(𝑛−1)𝑥𝑛−2+𝑎2(𝑛−2)𝑥𝑛−3+…+𝑎𝑛−1=nxn−1+a(n−1)xn−2+a2(n−2)xn−3+…+an−1

    Question 6

    Find the derivative of each given function.

    (i) (𝑥−𝑎)(𝑥−𝑏)(x−a)(x−b)

    Using the product rule:

    𝑑𝑑𝑥[(𝑥−𝑎)(𝑥−𝑏)]=(𝑥−𝑎)𝑑𝑑𝑥(𝑥−𝑏)+(𝑥−𝑏)𝑑𝑑𝑥(𝑥−𝑎)dxd​[(x−a)(x−b)]=(x−a)dxd​(x−b)+(x−b)dxd​(x−a)=(𝑥−𝑎)⋅1+(𝑥−𝑏)⋅1=(x−a)⋅1+(x−b)⋅1=𝑥−𝑎+𝑥−𝑏=x−a+x−b=2𝑥−(𝑎+𝑏)=2x−(a+b)

    (ii) (𝑥2+𝑏)2(x2+b)2

    Using the chain rule:

    𝑑𝑑𝑥[(𝑥2+𝑏)2]=2(𝑥2+𝑏)⋅𝑑𝑑𝑥(𝑥2+𝑏)dxd​[(x2+b)2]=2(x2+b)⋅dxd​(x2+b)=2(𝑥2+𝑏)⋅2𝑥=2(x2+b)⋅2x=4𝑥(𝑥2+𝑏)=4x(x2+b)

    (iii) 𝑥−𝑎𝑥−𝑏x−bx−a​

    Using the quotient rule:

    𝑑𝑑𝑥[𝑥−𝑎𝑥−𝑏]=(𝑥−𝑏)⋅𝑑𝑑𝑥(𝑥−𝑎)−(𝑥−𝑎)⋅𝑑𝑑𝑥(𝑥−𝑏)(𝑥−𝑏)2dxd​[x−bx−a​]=(x−b)2(x−b)⋅dxd​(x−a)−(x−a)⋅dxd​(x−b)​=(𝑥−𝑏)⋅1−(𝑥−𝑎)⋅1(𝑥−𝑏)2=(x−b)2(x−b)⋅1−(x−a)⋅1​=𝑥−𝑏−(𝑥−𝑎)(𝑥−𝑏)2=(x−b)2x−b−(x−a)​=−𝑏+𝑎(𝑥−𝑏)2=(x−b)2−b+a​=𝑎−𝑏(𝑥−𝑏)2=(x−b)2a−b​

    Question 7

    Find the derivative of 𝑥𝑛−𝑎𝑛𝑥−𝑎x−axn−an​ for some constant 𝑎a.

    This is a typical limit definition of the derivative at 𝑥=𝑎x=a. We simplify first using L'Hôpital's rule because it results in the indeterminate form 0000​.

    Using L'Hôpital's Rule:

    𝑑𝑑𝑥[𝑥𝑛−𝑎𝑛𝑥−𝑎]=lim⁡𝑥→𝑎𝑑𝑑𝑥(𝑥𝑛−𝑎𝑛)𝑑𝑑𝑥(𝑥−𝑎)dxd​[x−axn−an​]=limdxd​(x−a)dxd​(xn−an)​=lim⁡𝑥→𝑎𝑛𝑥𝑛−11=lim1nxn−1​=𝑛𝑎𝑛−1=nan−1

    Question 8

    Find the derivative of each given function.

    (i) 2𝑥−342x−43​

    Solution:

    𝑑𝑑𝑥(2𝑥−34)=2dxd​(2x−43​)=2

    (ii) (5𝑥2+3𝑥−1)(𝑥−1)(5x2+3x−1)(x−1)

    Using the product rule:

    𝑑𝑑𝑥[(5𝑥2+3𝑥−1)(𝑥−1)]=(5𝑥2+3𝑥−1)⋅𝑑𝑑𝑥(𝑥−1)+(𝑥−1)⋅𝑑𝑑𝑥(5𝑥2+3𝑥−1)dxd​[(5x2+3x−1)(x−1)]=(5x2+3x−1)⋅dxd​(x−1)+(x−1)⋅dxd​(5x2+3x−1)=(5𝑥2+3𝑥−1)⋅1+(𝑥−1)⋅(10𝑥+3)=(5x2+3x−1)⋅1+(x−1)⋅(10x+3)=5𝑥2+3𝑥−1+(𝑥−1)(10𝑥+3)=5x2+3x−1+(x−1)(10x+3)=5𝑥2+3𝑥−1+10𝑥2+3𝑥−10𝑥−3=5x2+3x−1+10x2+3x−10x−3=15𝑥2+6𝑥−4=15x2+6x−4

    (iii) 𝑥−3(5+3𝑥)x−3(5+3x)

    Using the product rule:

    𝑑𝑑𝑥[𝑥−3(5+3𝑥)]=𝑥−3⋅𝑑𝑑𝑥(5+3𝑥)+(5+3𝑥)⋅𝑑𝑑𝑥(𝑥−3)dxd​[x−3(5+3x)]=x−3⋅dxd​(5+3x)+(5+3x)⋅dxd​(x−3)=𝑥−3⋅3+(5+3𝑥)⋅(−3𝑥−4)=x−3⋅3+(5+3x)⋅(−3x−4)=3𝑥−3−15𝑥−4−9𝑥−3=3x−3−15x−4−9x−3=−6𝑥−3−15𝑥−4=−6x−3−15x−4

    (iv) 𝑥5(3−6𝑥−9)x5(3−6x−9)

    Using the product rule:

    𝑑𝑑𝑥[𝑥5(3−6𝑥−9)]=𝑥5⋅𝑑𝑑𝑥(3−6𝑥−9)+(3−6𝑥−9)⋅𝑑𝑑𝑥(𝑥5)dxd​[x5(3−6x−9)]=x5⋅dxd​(3−6x−9)+(3−6x−9)⋅dxd​(x5)=𝑥5⋅(−6)⋅(−9𝑥−10)+(3−6𝑥−9)⋅(5𝑥4)=x5⋅(−6)⋅(−9x−10)+(3−6x−9)⋅(5x4)=54𝑥−5+15𝑥4−30𝑥−5=54x−5+15x4−30x−5=15𝑥4+24𝑥−5=15x4+24x−5

12. 12.Exercise Question

    Let's solve the derivatives of each given function step-by-step.

    1. (𝑥+𝑎)(x+a)

    𝑓(𝑥)=𝑥+𝑎f(x)=x+a The derivative of a linear function is the coefficient of 𝑥x: 𝑓′(𝑥)=1f′(x)=1

    2. (𝑝𝑥+𝑞)(px+q)

    𝑓(𝑥)=𝑝𝑥+𝑞f(x)=px+q The derivative of a linear function is the coefficient of 𝑥x: 𝑓′(𝑥)=𝑝f′(x)=p

    3. −𝑟+𝑠𝑥x−r+s​

    Rewrite the function: 𝑓(𝑥)=(−𝑟+𝑠)𝑥−1f(x)=(−r+s)x−1 Using the power rule: 𝑓′(𝑥)=(−𝑟+𝑠)(−1)𝑥−2=−(𝑟−𝑠)𝑥2f′(x)=(−r+s)(−1)x−2=x2−(r−s)​

    4. (𝑎𝑥+𝑏)(𝑐𝑥+𝑑)2(cx+d)2(ax+b)​

    Using the quotient rule (𝑢𝑣)′=𝑢′𝑣−𝑢𝑣′𝑣2(vu​)′=v2u′v−uv′​: 𝑢=𝑎𝑥+𝑏⇒𝑢′=𝑎u=ax+b⇒u′=a 𝑣=(𝑐𝑥+𝑑)2⇒𝑣′=2(𝑐𝑥+𝑑)𝑐v=(cx+d)2⇒v′=2(cx+d)c

    𝑓′(𝑥)=𝑎(𝑐𝑥+𝑑)2−(𝑎𝑥+𝑏)2𝑐(𝑐𝑥+𝑑)(𝑐𝑥+𝑑)4=𝑎(𝑐𝑥+𝑑)−2𝑐(𝑎𝑥+𝑏)(𝑐𝑥+𝑑)3f′(x)=(cx+d)4a(cx+d)2−(ax+b)2c(cx+d)​=(cx+d)3a(cx+d)−2c(ax+b)​

    5. 𝑎𝑥+𝑏𝑐𝑥+𝑑cx+dax+b​

    Using the quotient rule: 𝑢=𝑎𝑥+𝑏⇒𝑢′=𝑎u=ax+b⇒u′=a 𝑣=𝑐𝑥+𝑑⇒𝑣′=𝑐v=cx+d⇒v′=c

    𝑓′(𝑥)=𝑎(𝑐𝑥+𝑑)−(𝑎𝑥+𝑏)𝑐(𝑐𝑥+𝑑)2=𝑎𝑐𝑥+𝑎𝑑−𝑎𝑐𝑥−𝑏𝑐(𝑐𝑥+𝑑)2=𝑎𝑑−𝑏𝑐(𝑐𝑥+𝑑)2f′(x)=(cx+d)2a(cx+d)−(ax+b)c​=(cx+d)2acx+ad−acx−bc​=(cx+d)2ad−bc​

    6. 1𝑥+11−1𝑥1−x1​x1​+1​

    Simplify the function first: 𝑓(𝑥)=1+𝑥𝑥−1f(x)=x−11+x​

    Using the quotient rule: 𝑢=1+𝑥⇒𝑢′=1u=1+x⇒u′=1 𝑣=𝑥−1⇒𝑣′=1v=x−1⇒v′=1

    𝑓′(𝑥)=(1)(𝑥−1)−(1+𝑥)(1)(𝑥−1)2=𝑥−1−1−𝑥(𝑥−1)2=−2(𝑥−1)2f′(x)=(x−1)2(1)(x−1)−(1+x)(1)​=(x−1)2x−1−1−x​=(x−1)2−2​

    7. 1𝑎𝑥2+𝑏𝑥+𝑐ax2+bx+c1​

    Using the chain rule 𝑑𝑑𝑥[𝑔(ℎ(𝑥))]=𝑔′(ℎ(𝑥))ℎ′(𝑥)dxd​[g(h(x))]=g′(h(x))h′(x): 𝑓(𝑥)=(𝑎𝑥2+𝑏𝑥+𝑐)−1f(x)=(ax2+bx+c)−1 𝑔(ℎ)=ℎ−1⇒𝑔′(ℎ)=−ℎ−2g(h)=h−1⇒g′(h)=−h−2 ℎ(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐⇒ℎ′(𝑥)=2𝑎𝑥+𝑏h(x)=ax2+bx+c⇒h′(x)=2ax+b

    𝑓′(𝑥)=−(𝑎𝑥2+𝑏𝑥+𝑐)−2(2𝑎𝑥+𝑏)=−2𝑎𝑥+𝑏(𝑎𝑥2+𝑏𝑥+𝑐)2f′(x)=−(ax2+bx+c)−2(2ax+b)=−(ax2+bx+c)22ax+b​

    8. 𝑎𝑥+𝑏𝑝𝑥2+𝑞𝑥+𝑟px2+qx+rax+b​

    Using the quotient rule: 𝑢=𝑎𝑥+𝑏⇒𝑢′=𝑎u=ax+b⇒u′=a 𝑣=𝑝𝑥2+𝑞𝑥+𝑟⇒𝑣′=2𝑝𝑥+𝑞v=px2+qx+r⇒v′=2px+q

    𝑓′(𝑥)=𝑎(𝑝𝑥2+𝑞𝑥+𝑟)−(𝑎𝑥+𝑏)(2𝑝𝑥+𝑞)(𝑝𝑥2+𝑞𝑥+𝑟)2=𝑎𝑝𝑥2+𝑎𝑞𝑥+𝑎𝑟−2𝑎𝑝𝑥2−𝑎𝑞𝑥−2𝑏𝑝𝑥−𝑏𝑞(𝑝𝑥2+𝑞𝑥+𝑟)2f′(x)=(px2+qx+r)2a(px2+qx+r)−(ax+b)(2px+q)​=(px2+qx+r)2apx2+aqx+ar−2apx2−aqx−2bpx−bq​

    𝑓′(𝑥)=𝑎𝑝𝑥2−2𝑎𝑝𝑥2+𝑎𝑞𝑥−𝑎𝑞𝑥+𝑎𝑟−2𝑏𝑝𝑥−𝑏𝑞(𝑝𝑥2+𝑞𝑥+𝑟)2=−𝑎𝑝𝑥2+𝑎𝑟−2𝑏𝑝𝑥−𝑏𝑞(𝑝𝑥2+𝑞𝑥+𝑟)2f′(x)=(px2+qx+r)2apx2−2apx2+aqx−aqx+ar−2bpx−bq​=(px2+qx+r)2−apx2+ar−2bpx−bq​

    9. 𝑝𝑥2+𝑞𝑥+𝑟𝑎𝑥+𝑏ax+bpx2+qx+r​

    Using the quotient rule: 𝑢=𝑝𝑥2+𝑞𝑥+𝑟⇒𝑢′=2𝑝𝑥+𝑞u=px2+qx+r⇒u′=2px+q 𝑣=𝑎𝑥+𝑏⇒𝑣′=𝑎v=ax+b⇒v′=a

    𝑓′(𝑥)=(2𝑝𝑥+𝑞)(𝑎𝑥+𝑏)−(𝑝𝑥2+𝑞𝑥+𝑟)𝑎(𝑎𝑥+𝑏)2f′(x)=(ax+b)2(2px+q)(ax+b)−(px2+qx+r)a​

    10. 𝑎𝑥4+𝑏𝑥2+cos⁡𝑥x4a​+x2b​+cosx

    Rewrite the function: 𝑓(𝑥)=𝑎𝑥−4+𝑏𝑥−2+cos⁡𝑥f(x)=ax−4+bx−2+cosx Using the power rule and trigonometric derivative: 𝑓′(𝑥)=−4𝑎𝑥−5−2𝑏𝑥−3−sin⁡𝑥f′(x)=−4ax−5−2bx−3−sinx

    11. 4𝑥2−24x2−2​

    Using the chain rule: 𝑓(𝑥)=4(𝑥2−2)1/2f(x)=4(x2−2)1/2 𝑔(ℎ)=ℎ1/2⇒𝑔′(ℎ)=12ℎ−1/2g(h)=h1/2⇒g′(h)=21​h−1/2 ℎ(𝑥)=𝑥2−2⇒ℎ′(𝑥)=2𝑥h(x)=x2−2⇒h′(x)=2x

    𝑓′(𝑥)=4⋅12(𝑥2−2)−1/2⋅2𝑥=4𝑥𝑥2−2f′(x)=4⋅21​(x2−2)−1/2⋅2x=x2−2​4x​

    12. (𝑎𝑥+𝑏)𝑛(ax+b)n

    Using the chain rule: 𝑓(𝑥)=(𝑎𝑥+𝑏)𝑛f(x)=(ax+b)n 𝑔(ℎ)=ℎ𝑛⇒𝑔′(ℎ)=𝑛ℎ𝑛−1g(h)=hn⇒g′(h)=nhn−1 ℎ(𝑥)=𝑎𝑥+𝑏⇒ℎ′(𝑥)=𝑎h(x)=ax+b⇒h′(x)=a

    𝑓′(𝑥)=𝑛(𝑎𝑥+𝑏)𝑛−1⋅𝑎=𝑛𝑎(𝑎𝑥+𝑏)𝑛−1f′(x)=n(ax+b)n−1⋅a=na(ax+b)n−1

    13. (𝑎𝑥+𝑏)𝑛(𝑐𝑥+𝑑)𝑚(ax+b)n(cx+d)m

    Using the product rule: 𝑢=(𝑎𝑥+𝑏)𝑛⇒𝑢′=𝑛𝑎(𝑎𝑥+𝑏)𝑛−1u=(ax+b)n⇒u′=na(ax+b)n−1 𝑣=(𝑐𝑥+𝑑)𝑚⇒𝑣′=𝑚𝑐(𝑐𝑥+𝑑)𝑚−1v=(cx+d)m⇒v′=mc(cx+d)m−1

    𝑓′(𝑥)=(𝑎𝑥+𝑏)𝑛⋅𝑚𝑐(𝑐𝑥+𝑑)𝑚−1+(𝑐𝑥+𝑑)𝑚⋅𝑛𝑎(𝑎𝑥+𝑏)𝑛−1f′(x)=(ax+b)n⋅mc(cx+d)m−1+(cx+d)m⋅na(ax+b)n−1

    14. sin⁡(𝑥+𝑎)sin(x+a)

    Using the chain rule: 𝑓(𝑥)=sin⁡(𝑥+𝑎)f(x)=sin(x+a) 𝑔(ℎ)=sin⁡ℎ⇒𝑔′(ℎ)=cos⁡ℎg(h)=sinh⇒g′(h)=cosh ℎ(𝑥)=𝑥+𝑎⇒ℎ′(𝑥)=1h(x)=x+a⇒h′(x)=1

    𝑓′(𝑥)=cos⁡(𝑥+𝑎)f′(x)=cos(x+a)

    15. csc⁡𝑥cot⁡𝑥cscxcotx

    Using the product rule: 𝑢=csc⁡𝑥⇒𝑢′=−csc⁡𝑥cot⁡𝑥u=cscx⇒u′=−cscxcotx 𝑣=cot⁡𝑥⇒𝑣′=−csc⁡2𝑥v=cotx⇒v′=−csc2x

    𝑓′(𝑥)=csc⁡𝑥⋅(−csc⁡2𝑥)+cot⁡𝑥⋅(−csc⁡𝑥cot⁡𝑥)=−csc⁡3𝑥−csc⁡𝑥cot⁡2𝑥f′(x)=cscx⋅(−csc2x)+cotx⋅(−cscxcotx)=−csc3x−cscxcot2x

    16. cos⁡𝑥1+sin⁡𝑥1+sinxcosx​

    Using the quotient rule: 𝑢=cos⁡𝑥⇒𝑢′=−sin⁡𝑥u=cosx⇒u′=−sinx 𝑣=1+sin⁡𝑥⇒𝑣′=cos⁡𝑥v=1+sinx⇒v′=cosx

    𝑓′(𝑥)=(−sin⁡𝑥)(1+sin⁡𝑥)−(cos⁡𝑥)(cos⁡𝑥)(1+sin⁡𝑥)2=−sin⁡𝑥−sin⁡2𝑥−cos⁡2𝑥(1+sin⁡𝑥)2=−sin⁡𝑥−1(1+sin⁡𝑥)2=−1−sin⁡𝑥(1+sin⁡𝑥)2f′(x)=(1+sinx)2(−sinx)(1+sinx)−(cosx)(cosx)​=(1+sinx)2−sinx−sin2x−cos2x​=(1+sinx)2−sinx−1​=(1+sinx)2−1−sinx​