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Chapter 1: Precalculus Review & Limits
Key Concepts
 Functions and Graphs
 Polynomial, Rational, Exponential, and Logarithmic Functions
 Trigonometric Functions
 Inverse Functions
 Limits and Continuity

Step-by-Step Worked Examples
Example 1: Evaluate the limit

lim⁡x→3(2x2−5x+1)\lim_{x \to 3} (2x^2 - 5x + 1)limx→3(2x2−5x+1)

Solution:

1. Substitute x=3x = 3x=3 directly (polynomial limits can be evaluated directly).
2. Calculate: 2(3)2−5(3)+1=18−15+1=42(3)^2 - 5(3) + 1 = 18 - 15 + 1 =
42(3)2−5(3)+1=18−15+1=4. Answer: 4

Example 2: Limit of a rational function

lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2x−2x2−4

Solution:

1. Factor numerator: x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)x2−4=(x−2)(x+2).
2. Simplify: (x−2)(x+2)x−2=x+2\frac{(x-2)(x+2)}{x-2} = x + 2x−2(x−2)(x+2)=x+2.
3. Substitute x=2x = 2x=2: 2+2=42 + 2 = 42+2=4. Answer: 4

Example 3: Continuity Check

Determine if f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x-1}f(x)=x−1x2−1 is continuous at x=1x =
1x=1.

Solution:

1. Factor numerator: x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1)x2−1=(x−1)(x+1).
2. Simplify function: f(x)=x+1f(x) = x + 1f(x)=x+1, for x≠1x \
3. Find lim⁡x→1f(x)=2\lim_{x \to 1} f(x) = 2limx→1f(x)=2.
4. Function not defined at x = 1"), so discontinuous. **Answer:** Discontinuous at \(x = 1.

,Chapter 2: Derivatives
Key Concepts
 Definition of Derivative
 Power Rule, Product Rule, Quotient Rule
 Chain Rule
 Implicit Differentiation
 Higher Order Derivatives
 Derivatives of Trigonometric, Exponential, and Logarithmic Functions

Step-by-Step Worked Examples
Example 1: Basic derivative using power rule

Find ddx(5x4−3x2+7)\frac{d}{dx} (5x^4 - 3x^2 + 7)dxd(5x4−3x2+7).

Solution:

1. Apply power rule term by term: ddx[xn]=nxn−1\frac{d}{dx}[x^n] = nx^{n-1}dxd
[xn]=nxn−1.
2. Calculate:
o ddx[5x4]=20x3\frac{d}{dx}[5x^4] = 20x^3dxd[5x4]=20x3
o ddx[−3x2]=−6x\frac{d}{dx}[-3x^2] = -6xdxd[−3x2]=−6x
o ddx[7]=0\frac{d}{dx}[7] = 0dxd[7]=0
3. Combine: 20x3−6x20x^3 - 6x20x3−6x Answer: 20x3−6x20x^3 - 6x20x3−6x

Example 2: Product Rule

Find ddx[(x2+1)(sin⁡x)]\frac{d}{dx}[(x^2 + 1)(\sin x)]dxd[(x2+1)(sinx)].

Solution:

1. Apply product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′.
2. Let u=x2+1u = x^2 + 1u=x2+1, v=sin⁡xv = \sin xv=sinx.
o u′=2xu' = 2xu′=2x, v′=cos⁡xv' = \cos xv′=cosx
3. Compute: u′v+uv′=(2x)(sin⁡x)+(x2+1)(cos⁡x)u'v + uv' = (2x)(\sin x) + (x^2 + 1)(\cos
x)u′v+uv′=(2x)(sinx)+(x2+1)(cosx) Answer: 2xsin⁡x+(x2+1)cos⁡x2x\sin x + (x^2 +
1)\cos x2xsinx+(x2+1)cosx

Example 3: Chain Rule

, Find ddx[ln⁡(3x2+2)]\frac{d}{dx}[\ln(3x^2 + 2)]dxd[ln(3x2+2)].

Solution:

1. Recall ddx[ln⁡f(x)]=f′(x)f(x)\frac{d}{dx}[\ln f(x)] = \frac{f'(x)}{f(x)}dxd
[lnf(x)]=f(x)f′(x).
2. Compute f(x)=3x2+2f(x) = 3x^2 + 2f(x)=3x2+2, f′(x)=6xf'(x) = 6xf′(x)=6x.
3. Apply formula: 6x3x2+2\frac{6x}{3x^2 + 2}3x2+26x Answer: 6x3x2+2\frac{6x}{3x^2
+ 2}3x2+26x

Basic Problems

Problem 1: Find f′(x)f'(x)f′(x) if f(x)=x3.f(x) = x^3.f(x)=x3.

 Solution: Power rule: f′(x)=3x2.f'(x) = 3x^2.f′(x)=3x2.
 Answer: 3x23x^23x2

Problem 2: Differentiate f(x)=5x2−7x+4.f(x) = 5x^2 - 7x + 4.f(x)=5x2−7x+4.

 Solution: Term by term: f′(x)=10x−7.f'(x) = 10x - 7.f′(x)=10x−7.
 Answer: 10x−710x - 710x−7

Problem 3: Differentiate f(x)=sin⁡x.f(x) = \sin x.f(x)=sinx.

 Solution: Derivative: f′(x)=cos⁡x.f'(x) = \cos x.f′(x)=cosx.
 Answer: cos⁡x\cos xcosx

Problem 4: Differentiate f(x)=ex.f(x) = e^x.f(x)=ex.

 Solution: Derivative: ex.e^x.ex.
 Answer: exe^xex

Problem 5: Differentiate f(x)=ln⁡x.f(x) = \ln x.f(x)=lnx.

 Solution: Derivative: 1/x.1/x.1/x.
 Answer: 1/x1/x1/x



Intermediate Problems

Problem 6: Differentiate f(x)=x2ex.f(x) = x^2 e^x.f(x)=x2ex.

 Solution: Product rule: (2x)(ex)+(x2)(ex)=(2x+x2)ex.(2x)(e^x) + (x^2)(e^x) = (2x +
x^2)e^x.(2x)(ex)+(x2)(ex)=(2x+x2)ex.
 Answer: (x2+2x)ex(x^2 + 2x)e^x(x2+2x)ex