Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　２．３

Tｈｅ　Pｒｏｄｕｃｔ　ａｎｄ　Qｕｏｔｉｅｎｔ　Rｕｌｅｓ　Fｏｒ　Dｅｒｉｖａｔｉｖｅｓ　Oｆ　Fｕｎｃｔｉｏｎｓ

I ｎｔｒｏｄｕｃｔｉｏｎ.

• 0:00. Introduction to the lesson on the product rule and the quotient rule for derivatives.
• 0:19. Brief review of the lesson topics: product rule and quotient rule.

Ｐｒｏｄｕｃｔ ｒｕｌｅ

• 0:41. Presentation of the scenario: Is it possible to derive the product of two functions?
  • 0:52. 🧩 Example – with basic functions: $𝒇(𝒙)=𝒙^{2}$ and $𝓰(𝒙)=𝒙^{3}$. 📷 – [📷image]
    • Posing the main question: Is $\dfrac{d}{dx}\big[𝒇(𝒙)\cdot 𝓰(𝒙)\big]=\dfrac{d}{dx}\big[𝒇(𝒙)\big]\cdot \dfrac{d}{dx}\big[𝓰(𝒙)\big]$?
  • 1:47. Analysis of the question through the given Example.
    • It is observed that $𝒇(𝒙)\cdot 𝓰(𝒙)=𝒙^{5}$.
      • Calculating $\dfrac{d}{dx}\big(𝒙^{5}\big)=5𝒙^{4}$.
    • Calculating separately $\dfrac{d}{dx}\big[𝒇(𝒙)\big]\cdot \dfrac{d}{dx}\big[𝓰(𝒙)\big]$.
      • A different answer is obtained: $6𝒙^{3}$.
    • Conclusion: Derivatives cannot be separated by multiplication.
• 4:48. Need for a rule to derive products of functions.
  • 5:00. Introduction to the concept of the product rule.
• 5:21. Explanation of the product rule.
  • 6:05. Interpretation in words of the product rule.
  • Formula of the product rule:
    • $\dfrac{d}{dx}\big[𝒇(𝒙)\cdot 𝓰(𝒙)\big]=\dfrac{d}{dx}\big[𝒇(𝒙)\big]\cdot 𝓰(𝒙)+𝒇(𝒙)\cdot \dfrac{d}{dx}\big[𝓰(𝒙)\big]$.
• 7:13. Verification of the product rule using the initial Example. 📷 – [📷image]
  • Emphasis on the importance of the notation $\dfrac{d}{dx}$ to indicate the function to be derived.
    • $\dfrac{d}{dx}\big[𝒇(𝒙)\cdot 𝓰(𝒙)\big]=𝒇'(𝒙)\cdot 𝓰(𝒙)+𝒇(𝒙)\cdot 𝓰'(𝒙)$
    • $\dfrac{d}{dx}\big[𝒙^{2}\cdot 𝒙^{3}\big]=\dfrac{d}{dx}\big[𝒙^{2}\big]\cdot 𝒙^{3}+𝒙^{2}\cdot \dfrac{d}{dx}\big[𝒙^{3}\big]$
    • $2𝒙\cdot 𝒙^{3}+𝒙^{2}\cdot 3𝒙^{2}$
    • $2𝒙^{4}+3𝒙^{4}=5𝒙^{4}$
  • Mention of the need for a formal proof to generalize the rule.

Ｐｒｅｌｉｍｉｎａｒｙ　ｃｏｎｓｉｄｅｒａｔｉｏｎｓ　ｆｏｒ　ｔｈｅ　ｐｒｏｄｕｃｔ　ｒｕｌｅ：　ｅｖａｌｕａｔｉｎｇ　ｄｉｓｔｒｉｂｕｔｉｏｎ.

• 10:34. Presentation of an example where distribution is possible.
  • 🧩 Example –: $(𝒙^{2}-1)\cdot(3𝒙^{4}-2𝒙)$
    • 11:09. Anticipation of scenarios where distribution will n̲o̲t̲ be feasible.
      • 11:23. 🧩 Example – with a high exponent, where distribution would be tedious.
        • $(𝒙^{2}-1)\cdot(3𝒙^{4}-2𝒙)^{4}$
        • This is why the project rule become very important
    • 12:19. 🧩 Example – A Constant in the Product Rule 📷 – [📷image]
      • Why is the product rule not used here? Because the constant's derivative is zero, so it contributes nothing.
    • 14:10. Two ways to solve $𝐲'$ from $𝐲=(𝒙^{2}-1)(3𝒙^{4}+2𝒙)$ 📷 – [📷image]
      • Applying distribution
        • Expand:
          • $𝐲=3𝒙^{6}+2𝒙^{3}-3𝒙^{4}-2𝒙$
          • $𝐲=3𝒙^{6}-3𝒙^{4}+2𝒙^{3}-2𝒙$
        • Derivative (power rule):
          • $\dfrac{d𝐲}{d𝒙}=18𝒙^{5}-12𝒙^{3}+6𝒙^{2}-2$
      • 15:15. Applying the product rule Given:
        • Product rule Setup
          • Aply formula:
            • $\\dfrac{d𝐲}{d𝒙}=\\dfrac{d}{d𝒙}\big[(𝒙^{2}-1)(3𝒙^{4}+2𝒙)\big]=\\dfrac{d}{d𝒙}\big[𝒙^{2}-1\big]\cdot(3𝒙^{4}+2𝒙)+(𝒙^{2}-1)\cdot \\dfrac{d}{d𝒙}\big[3𝒙^{4}+2𝒙\big]$
          • Differentiate each part:
            • $=(2𝒙)(3𝒙^{4}+2𝒙)+(𝒙^{2}-1)(12𝒙^{3}+2)$
          • Expand both products:
            • $=6𝒙^{5}+4𝒙^{2}+12𝒙^{5}+2𝒙^{2}-12𝒙^{3}-2$
          • Combine like terms:
            • $\\dfrac{d𝐲}{d𝒙}=18𝒙^{5}-12𝒙^{3}+6𝒙^{2}-2$

Ａｐｐｌｉｃａｔｉｏｎ　ｏｆ　ｔｈｅ　ｐｒｏｄｕｃｔ　ｒｕｌｅ

• 19:36. You need to understand the product rule first because it's one of the fundamental tools for differentiating expressions where two functions are multiplied: $\big(𝒇(𝒙)\cdot 𝓰(𝒙)\big)'=𝒇'(𝒙)\cdot 𝓰(𝒙)+𝒇(𝒙)\cdot 𝓰'(𝒙)$. Later on, you'll often encounter problems where this rule appears in combination with the chain rule: $\big(𝒇(𝓰(𝒙))\big)'=𝒇'\big(𝓰(𝒙)\big)\cdot 𝓰'(𝒙)$ or with the quotient rule: $\big(\dfrac{𝒇(𝒙)}{𝓰(𝒙)}\big)'=\dfrac{𝒇'(𝒙)\cdot 𝓰(𝒙)-𝒇(𝒙)\cdot 𝓰'(𝒙)}{\big(𝓰(𝒙)\big)^{2}}$. Understanding the product rule early will make it easier to handle these more complex derivative rules as they often work together.
• 20:10. 🧩 Example – $(𝒙^{2}+1)\cdot \sqrt{𝒙}$ 📷 – [📷image]
  • Apply the product rule:
    •   $𝒇'(𝒙)=\dfrac{d}{d𝒙}[1+𝒙^{2}]\cdot 𝒙^{1/2}+(1+𝒙^{2})\cdot \dfrac{d}{d𝒙}\big[𝒙^{1/2}\big]$
    •    $𝒇'(𝒙)=2𝒙\cdot 𝒙^{1/2}+(1+𝒙^{2})\cdot \dfrac{1}{2}\cdot 𝒙^{-1/2}$
  • Distribute:
    •    $𝒇'(𝒙)=2𝒙\cdot 𝒙^{1/2}+\dfrac{1}{2}\cdot 𝒙^{-1/2}+\dfrac{1}{2}\cdot 𝒙^{2}\cdot 𝒙^{-1/2}$
    •    $𝒇'(𝒙)=2\cdot 𝒙^{3/2}+\dfrac{1}{2}\cdot 𝒙^{-1/2}+\dfrac{1}{2}\cdot 𝒙^{3/2}$
    •    $𝒇'(𝒙)=\dfrac{5}{2}\cdot 𝒙^{3/2}+\dfrac{1}{2}\cdot 𝒙^{-1/2}$
• 32:10. 🧩 Example – Find $𝓰'(2)$ ; $𝓰(𝒙)=(𝒙^{2}+1)\cdot 𝒇(𝒙)$ where $𝒇(2)=3$ and $𝒇'(2)=1$ 📷 – [📷image]
  • Apply the product rule:
    • $𝓰'(𝒙)=\dfrac{d}{d𝒙}\big[(𝒙^{2}+1)\big]\cdot 𝒇(𝒙)+(𝒙^{2}+1)\cdot \dfrac{d}{d𝒙}\big[𝒇(𝒙)\big]$
    • $𝓰'(𝒙)=2𝒙\cdot 𝒇(𝒙)+(𝒙^{2}+1)\cdot 𝒇'(𝒙)$
  • Substitute $𝒙=2$:
    • $𝓰'(2)=2\cdot 2\cdot 𝒇(2)+(2^{2}+1)\cdot 𝒇'(2)$
    • $𝓰'(2)=4\cdot 3+5\cdot(-1)$
    • $𝓰'(2)=12-5=7$

Ｑｕｏｔｉｅｎｔ　ｒｕｌｅ

• 39:10. Posing the question: Is it possible to derive the quotient of two functions?
  • Dismissing the possibility of deriving the numerator and the denominator separately.
• 40:00. Introduction to the quotient rule.
  • 40:50. Presentation of the formula for the quotient rule. 📷 – [📷image]
    • $\dfrac{d}{d𝒙}\Big[\dfrac{𝒇(𝒙)}{𝓰(𝒙)}\Big]=\dfrac{𝓰(𝒙)\cdot 𝒇'(𝒙)-𝒇(𝒙)\cdot 𝓰'(𝒙)}{\big(𝓰(𝒙)\big)^{2}}$
    • Likewise $\dfrac{d}{d𝒙}\Big[\dfrac{𝒇(𝒙)}{𝓰(𝒙)}\Big]=\dfrac{𝒇'(𝒙)\cdot 𝓰(𝒙)-𝒇(𝒙)\cdot 𝓰'(𝒙)}{\big(𝓰(𝒙)\big)^{2}}$
  • 42:20. Other notation
    • $\dfrac{𝓰(𝒙)\cdot \dfrac{d}{d𝒙}\big[𝒇(𝒙)\big]-𝒇(𝒙)\cdot \dfrac{d}{d𝒙}\big[𝓰(𝒙)\big]}{\big(𝓰(𝒙)\big)^{2}}$
    • Likewise $\dfrac{\dfrac{d}{d𝒙}\big[𝒇(𝒙)\big]\cdot 𝓰(𝒙)-𝒇(𝒙)\cdot \dfrac{d}{d𝒙}\big[𝓰(𝒙)\big]}{\big(𝓰(𝒙)\big)^{2}}$

Ａｐｐｌｉｃａｔｉｏｎ　　ｅｘａｍｐｌｅｓ　ｏｆ　ｔｈｅ　ｑｕｏｔｉｅｎｔ　ｒｕｌｅ

• 45:40. 🧩 Example –: $𝑦=\dfrac{𝒙^{3}-3𝒙^{2}-5}{2𝒙+5}$
  • An example where the quotient rule is necessary.
  • 46:10. Comparison with a case where prior simplification is possible.
    • $𝑦=\dfrac{𝒙^{3}-3𝒙^{2}-5}{2𝒙}$
      • Iinstead of directly using the quotient rule, you can split the fraction into simpler parts, combine exponents, and differentiate each piece separatel.
  • 48:00. Explanation of how to set up the quotient rule for the given example. 📷 – [📷image]
    • Apply the quotient rule:
      • $\dfrac{d𝑦}{d𝒙}=\dfrac{(2𝒙+5)\cdot \dfrac{d}{d𝒙}(𝒙^{3}-3𝒙^{2}-5)-(𝒙^{3}-3𝒙^{2}-5)\cdot \dfrac{d}{d𝒙}(2𝒙+5)}{(2𝒙+5)^{2}}$
    • Calculate derivatives:
      • $\dfrac{d}{d𝒙}(𝒙^{3}-3𝒙^{2}-5)=3𝒙^{2}-6𝒙$
      • $\dfrac{d}{d𝒙}(2𝒙+5)=2$
    • Substitute and expand:
      • Numerator: $(2𝒙+5)(3𝒙^{2}-6𝒙)-(𝒙^{3}-3𝒙^{2}-5)\cdot 2$
      • Denominator: $(2𝒙+5)^{2}$
    • Expand the numerator:
      • $6𝒙^{3}-12𝒙^{2}+15𝒙^{2}-30𝒙-2𝒙^{3}+6𝒙^{2}+10$
    • Simplify:
      • $4𝒙^{3}+9𝒙^{2}-30𝒙+10$
    • Final answer:
      • $\dfrac{d𝑦}{d𝒙}=\dfrac{4𝒙^{3}+9𝒙^{2}-30𝒙+10}{(2𝒙+5)^{2}}$
    • Emphasis on the importance of using parentheses for correct distribution.

Ｐｒｏｄｕｃｔ　ｒｕｌｅ　ｗｉｔｈｉｎ　ｔｈｅ　ｑｕｏｔｉｅｎｔ　ｒｕｌｅ

• 57:55. 🧩 Example – where the product rule and quotient rule are combined: $𝒇(𝒙)=\dfrac{(3𝒙-1)(𝒙^{2}+4)}{𝒙^{2}+2}$ 📷 – [📷image]
  • Explanation of the importance of identifying which rule has higher hierarchy in the problem.
• 58:20. Description of setting up the quotient rule for the given example, including the product rule within it.
  • 1:00:50. Emphasis on the importance of following the notation $\dfrac{d}{d𝒙}$ to perform derivatives correctly.
  • Apply the quotient rule:
    • $𝒇'(𝒙)=\dfrac{(𝒙^{2}+2)\cdot \dfrac{d}{d𝒙}\big[(3𝒙-1)(𝒙^{2}+4)\big]-(3𝒙-1)(𝒙^{2}+4)\cdot \dfrac{d}{d𝒙}\big[𝒙^{2}+2\big]}{(𝒙^{2}+2)^{2}}$
  • Calculate derivatives:
    • $\dfrac{d}{d𝒙}\big[(3𝒙-1)(𝒙^{2}+4)\big]=3(𝒙^{2}+4)+2𝒙(3𝒙-1)$
    • $\dfrac{d}{d𝒙}\big[𝒙^{2}+2\big]=2𝒙$
  • Substitute and expand:
    • Numerator: $(𝒙^{2}+2)\big[3(𝒙^{2}+4)+2𝒙(3𝒙-1)\big]-(3𝒙-1)(𝒙^{2}+4)\cdot 2𝒙$
    • Denominator: $(𝒙^{2}+2)^{2}$