Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　２．６

　Dｉｓｃｕｓｓｉｏｎ　Oｆ　Tｈｅ　Cｈａｉｎ　Rｕｌｅ　Fｏｒ　Dｅｒｉｖａｔｉｖｅｓ　Oｆ　Fｕｎｃｔｉｏｎｓ

Ｉｎｔｒｏｄｕｃｔｉｏｎ　ａｎｄ　ｒｅｌｅｖａｎｃｅ　ｏｆ　ｔｈｅ　ｃｈａｉｎ　ｒｕｌｅ

• 0:00. Introduction to the Chain Rule and its importance in calculus.
  • The Cha+in Rule as the last main differentiation rule.
  • The three fundamental differentiation rules: Product Rule, Quotient Rule, and Chain Rule.
  • The Chain Rule as a method for **deriving compositions of functions.**

Ｍｏｔｉｖａｔｉｏｎ　ａｎｄ　ｅｘａｍｐｌｅｓ　ｏｆ　ｔｈｅ　ｃｈａｉｎ　ｒｕｌｅ

• 1:19. Review of the concept of function compositions.
• 1:26. Example illustrating the need for the Chain Rule: derivative of $(3𝒙^{2}-4)^{100}$
  • Presentation of examples of derivatives that can be calculated without the chain rule:
    • $\dfrac{d}{d𝒙}\big[3𝒙^{2}\cdot 4\big]$
    • $\dfrac{d}{d𝒙}\big[(3𝒙^{2}\cdot 4)^{2}\big]$; you can apply the product rule or expand, then differentiate.
    • $\dfrac{d}{d𝒙}\big[(3𝒙^{2}\cdot 4)^{3}\big]$; still doable, but becomes tedious.
  • Presentation of examples of derivatives that are difficult to calculate _without the chain rule_:
    • $\dfrac{d}{d𝒙}\big[(3𝒙^{2}\cdot 4)^{100}\big]$
  • Difficulty of deriving the expression $(3𝒙^{2}-4)^{100}$ using traditional methods.
• 2:31. The Chain Rule as an efficient solution for this type of derivative.

Ｅｘｐｌａｎａｔｉｏｎ　ｏｆ　ｔｈｅ　ｃｈａｉｎ　ｒｕｌｅ　ｔｈｒｏｕｇｈ　ａｎ　ｅｘａｍｐｌｅ

• 3:06. Expression of $(3𝒙^{2}-4)^{100}$ as a ᴄ̳ᴏ̳ᴍ̳ᴩ̳ᴏ̳ꜱ̳ɪ̳ᴛ̳ɪ̳ᴏ̳ɴ̳ ̳ᴏ̳ꜰ̳ ̳ꜰ̳ᴜ̳ɴ̳ᴄ̳ᴛ̳ɪ̳ᴏ̳ɴ̳ꜱ̳. – [📷image]
  • Motivation: Recognizing compositions is essential before applying the chain rule.
• 3:40. Method to identify the ɪ̳ɴ̳ɴ̳ᴇ̳ʀ̳ ̳ᴀ̳ɴ̳ᴅ̳ ̳ᴏ̳ᴜ̳ᴛ̳ᴇ̳ʀ̳ ̳ꜰ̳ᴜ̳ɴ̳ᴄ̳ᴛ̳ɪ̳ᴏ̳ɴ̳ꜱ̳ ̳ɪ̳ɴ̳ ̳ᴀ̳ ̳ᴄ̳ᴏ̳ᴍ̳ᴩ̳ᴏ̳ꜱ̳ɪ̳ᴛ̳ɪ̳ᴏ̳ɴ̳.
  • 4:01. Definition of the "outer" and "inner" functions: $y=u^{100}$ and $u=3𝒙^{2}-4$.
  • 4:47. Verification: substituting $u$ into $y$ recovers the original function.
• 5:07. Calculation of $\dfrac{dy}{du}$ ($y=u^{100}$) and $\dfrac{du}{d𝒙}$ ($u=3𝒙^{2}-4$).
  • Objective: find $\dfrac{dy}{dx}$
  • $\dfrac{dy}{du}=100u^{99}$
  • $\dfrac{du}{d𝒙}=6𝒙$
  • Chain Rule in action: $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}=(\dfrac{dy}{\cancel{du}})\cdot(\dfrac{\cancel{du}}{dx})$
    • Notice how the intermediate variable $(u)$ "cancels" out, leaving the derivative in terms of $x$.
  • Mastering this identification process is crucial for differentiating any composite function efficiently.
• 7:50. Justification of Leibniz notation for the derivative.
  • The Chain Rule:
    • $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$
• 8:45. Application of the Chain Rule to the example $(3𝒙^{2}-4)^{100}$ – [📷image]
  • $\dfrac{d}{dx}\big[(3𝒙^{2}-4)^{100}\big]=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$
  • $\dfrac{d}{du}\big[u^{100}\big]\cdot\dfrac{d}{dx}\big[3𝒙^{2}-4\big]$
  • $100u^{99}\cdot 6𝒙$
  • $100u^{99}\cdot 6𝒙\Big|_{u=3𝒙^{2}-4}$
  • $100(3𝒙^{2}-4)^{99}\cdot 6𝒙$
  • $600𝒙(3𝒙^{2}-4)^{99}$
• 12:06. Simplification of the differentiation process using the Chain Rule.
  • Go for this step $\dfrac{d}{dx}\big[(3𝒙^{2}-4)^{100}\big]$ to this $100(3𝒙^{2}-4)^{99}\cdot 6𝒙$
  • The Chain Rule turns the derivative $\dfrac{d}{dx}\big[(3𝒙^{2}-4)^{100}\big]$ directly into $100(3𝒙^{2}-4)^{99}\cdot 6𝒙$, making the process systematic.
  • Remark:The Power Rule is actually a special case (corollary) of the Chain Rule, when the inner function is simply $x$.
    • In other words: if $y=x^{n}$, then $\dfrac{dy}{dx}=n x^{n-1}$, which fits the Chain Rule with inner function $u=x$.
    • The Chain Rule generalizes the Power Rule to any composition, not just powers of $x$.

Ｉｎｔｅｒｐｒｅｔａｔｉｏｎ　ａｎｄ　ｇｅｎｅｒａｌｉｚａｔｉｏｎ　ｏｆ　ｔｈｅ　Ｐｏｗｅｒ　Ｒｕｌｅ

• 13:20. Definition of the Generalized Power Rule. – [📷image]
• 14:05. Demonstration of the Chain Rule from the definition of compositions.
  • $\dfrac{d}{dx}\big[(𝒇(𝒙))^{n}\big]$
  • $y=u^{n},\;\;u=𝒇(𝒙)$
  • $\dfrac{dy}{du}=n\cdot u^{n-1};\;\; \dfrac{du}{dx}=\dfrac{d}{dx}\big[𝒇(𝒙)\big]$ $=\Big(\dfrac{dy}{du}\Big)\cdot\Big(\dfrac{du}{dx}\Big)$
  • $n\cdot u^{n-1}\cdot\dfrac{d}{dx}\big[𝒇(𝒙)\big]$
  • $n\cdot\big\{𝒇(𝒙)^{\,n-1}\cdot\dfrac{d}{dx}\big[𝒇(𝒙)\big]\big\}$
• 16:26. **General formula for the derivative of a function raised to a power**:
  • $\dfrac{d}{d𝒙}\big[(𝒇(𝒙))^{n}\big]=n\cdot\{𝒇(𝒙)^{\,n-1}\}\cdot\dfrac{d}{d𝒙}\big[𝒇(𝒙)\big]$
• 18:40. Statement in simple language of the Generalized Power Rule.
  • 1. Multiply by the exponent $n$.
  • 2. Keep the base function raised to $n-1$ (i.e., reduce the exponent by one).
  • 3. Multiply by the derivative of the base function.

Ｅｘａｍｐｌｅｓ　ａｎｄ　ａｐｐｌｉｃａｔｉｏｎｓ　ｏｆ　ｔｈｅ　ｃｈａｉｎ　ｒｕｌｅ

• 19:50. $y=(𝒙^{3}-2𝒙+38)^{4}$ find $y'$ – [📷image]
  • Importance of recognizing function compositions when applying the Chain Rule.
  • Common errors when applying the Generalized Power Rule (forgetting to subtract $1$ from the exponent).
  • Use of parentheses to indicate the multiplication of the inner derivative.
  • Relationship between the Generalized Power Rule and the simple Power Rule.

Ａｄｄｉｔｉｏｎａｌ　ｅｘａｍｐｌｅｓ：ｃｏｍｂｉｎａｔｉｏｎ　ｏｆ　ｄｉｆｆｅｒｅｎｔｉａｔｉｏｎ　ｒｕｌｅｓ

• 24:25. 🧩 Example –: $\dfrac{d}{d𝒙}\big[(𝒙)^{6}\big]$ – [📷image]
  • Could you do the general power rule?
• 26:20. 🧩 Example –: $\dfrac{d}{d𝒙}\big[(2𝒙-3)(𝒙^{2}-5)^{3}\big]$ – [📷image]
  • Example combining the Product Rule and the Generalized Power Rule.
  • Determining which rule to apply first: the Product Rule applies to the entire expression.
    • 27:28. $\dfrac{d}{d𝒙}\Big[\dfrac{(𝒙^{2}-1)^{2}}{(𝒙+4)}\Big]$
      • Identifying the necessary differentiation rules: Chain Rule (or Generalized Power Rule) and Quotient Rule.
      • Order of application: Quotient Rule first
    • 28:00. $\dfrac{d}{d𝒙}\Big[\Big(\dfrac{(𝒙^{2}-1)^{2}}{(𝒙+4)}\Big)^{5}\Big]$
      • Order of application: Generalized Power Rule, then Quotient Rule, and finally another Generalized Power Rule within the Quotient Rule.
    • 29:38. $\dfrac{d}{d𝒙}\big[(2𝒙-3)(𝒙^{2}-5)^{3}\big]$
      • Application of the Product Rule as the main rule
  • 29:38. Application of the Product Rule as the main rule
    
    
    • $y=(𝒙^{3}-2𝒙+38)^{4}$
    
    
    • $\dfrac{dy}{dx}=4(𝒙^{3}-2𝒙+38)^{3}\cdot\dfrac{d}{dx}\big[𝒙^{3}-2𝒙+38\big]$
    
    
    • $4(𝒙^{3}-2𝒙+38)^{3}\big(3𝒙^{2}-2\big)$
    
    
    • $4\big(3𝒙^{2}-2\big)(𝒙^{3}-2𝒙+38)^{3}$
    
    
    • $(12𝒙^{2}-8)(𝒙^{3}-2𝒙+38)^{3}$
• 39:05. 🧩 Example –: $\dfrac{d}{d𝒙}\big[\sqrt{\,5𝒙^{2}-1\,}\big]$ – [📷image]
  • Application of the Chain Rule to derivatives involving square roots.
  • Representing square roots as fractional exponents to facilitate differentiation.
  
  
  • $y=(5𝒙^{2}-1)^{1/2}$
  
  
  • $\dfrac{dy}{dx}=\dfrac{1}{2}(5𝒙^{2}-1)^{-1/2}\cdot\dfrac{d}{dx}\big[5𝒙^{2}-1\big]$
  
  
  • $\dfrac{1}{2}(5𝒙^{2}-1)^{-1/2}\cdot 10𝒙$
  
  
  • $\dfrac{5𝒙}{\sqrt{5𝒙^{2}-1}}$

Ｅｘｔｅｎｓｉｏｎ　ｏｆ　ｔｈｅ　ｃｈａｉｎ　ｒｕｌｅ　ｔｏ　ｔｒｉｇｏｎｏｍｅｔｒｉｃ　ｆｕｎｃｔｉｏｎｓ

• 43:08. Derivative of the function $\cos(𝒙^{4})$ using the Chain Rule. – [📷image]
  • Identifying the outer function (cosine) and the inner function $(𝒙^{4})$.
    • $y=\cos(u)$
    • $u=𝒙^{4}$
  • Applying the Chain Rule: derive the outer function, keep the inner function unchanged, and multiply by the derivative of the inner function.
    • $\dfrac{dy}{du}=-\sin(u)$
    • $\dfrac{du}{dx}=4𝒙^{3}$
    • $\dfrac{dy}{dx}=\Big(\dfrac{dy}{du}\Big)\cdot\Big(\dfrac{du}{dx}\Big)$ $=-\sin(u)\cdot 4𝒙^{3}$ $=-\sin(𝒙^{4})\cdot 4𝒙^{3}$ $=-4𝒙^{3}\sin(𝒙^{4})$
• 46:07. **Statement of the Chain Rule for trigonometric functions**:
  • $\dfrac{d}{dx}\big[𝒇(𝓰(𝒙))\big]=𝒇'\!\big(𝓰(𝒙)\big)\cdot 𝓰'(𝒙)$.

Ａｄｄｉｔｉｏｎａｌ　ｅｘａｍｐｌｅｓ　ａｎｄ　ｐｒａｃｔｉｃｅ　ｗｉｔｈ　ｔｈｅ　ｃｈａｉｎ　ｒｕｌｅ

• 48:22. Practice with examples combining the Chain Rule, Product Rule, and Quotient Rule.
• 49:09. 🧩 Example –: $\dfrac{d}{d𝒙}\big[\sin(4𝒙^{5})\big]$ – [📷image]
  • Example of a trigonometric function with a composite argument
  • Identifying that this is a Chain Rule.
  • Recognizing the composition of functions: the outer function is sine and the inner function is $4𝒙^{5}$.
  • Applying the Chain Rule: derive the outer function, keep the inner function unchanged, and multiply by the derivative of the inner function.
    • $\dfrac{dy}{dx}=\cos(4𝒙^{5})\cdot\dfrac{d}{dx}\big[4𝒙^{5}\big]$ $=20𝒙^{4}\cos(4𝒙^{5})$
• 52:15. 🧩 Example –: $\dfrac{d}{d𝒙}\big[\cos^{2}(𝒙^{4})\big]$ – [📷image]
  • **Recommendation**: analyze the structure of a problem before applying differentiation rules.
  • Distinction between the general Chain Rule and the Generalized Power Rule.
  • In this context, the Generalized Power Rule refers to the Chain Rule applied to a function raised to a power.
    
    
    • $\dfrac{d}{d𝒙}\big[\cos^{2}(𝒙^{4})\big]=\dfrac{d}{d𝒙}\big[(\cos(𝒙^{4}))^{2}\big]$
    
    
    • $2\cos(𝒙^{4})\cdot\dfrac{d}{dx}\big[\cos(𝒙^{4})\big]$
    
    
    • $2\cos(𝒙^{4})\cdot\big(-\sin(𝒙^{4})\cdot\dfrac{d}{dx}[𝒙^{4}]\big)$
    
    
    • $2\cos(𝒙^{4})\cdot\big(-\sin(𝒙^{4})\cdot 4𝒙^{3}\big)$
    
    
    • $-8𝒙^{3}\cos(𝒙^{4})\sin(𝒙^{4})$
• 1:01:08. 🧩 Example –: $\dfrac{d}{d𝒙}\big[\tan(3𝒙^{2}-2𝒙)\big]$ – [📷image]
  • If $\dfrac{d}{d𝒙}\big[\tan^{4}(3𝒙^{2}-2𝒙)\big]$ then first general power rule
  • Identifying the Chain Rule: the outer function is tangent and the inner function is $3𝒙^{2}-2𝒙$.
  • Applying the Chain Rule: derive the outer function, keep the inner function unchanged, and multiply by the derivative of the inner function.
    • $\dfrac{d}{dx}\big[\tan(3𝒙^{2}-2𝒙)\big]=\sec^{2}(3𝒙^{2}-2𝒙)\cdot\dfrac{d}{dx}[3𝒙^{2}-2𝒙]$
    • $\sec^{2}(3𝒙^{2}-2𝒙)\cdot(6𝒙-2)$
    • $(6𝒙-2)\sec^{2}(3𝒙^{2}-2𝒙)$
  • Common errors when applying the Chain Rule to trigonometric functions: forgetting to multiply by the derivative of the argument.
• 1:06:37. 🧩 Example –: $\dfrac{d}{d𝒙}\big[\sqrt{\,𝒙^{3}+\csc(𝒙^{3})\,}\big]$ – [📷image]
  • Simplifying the expression by moving the negative exponent to the denominator.
  • Identifying the Generalized Power Rule: the outer function is raising to the power of $-1/2$ and the inner function is the expression within the parentheses.
  • Applying the Generalized Power Rule: bring down the exponent, keep the internal expression unchanged, and multiply by the derivative of the internal expression.
  • To derive the internal function, it is necessary to apply the Sum Rule, Product Rule, and Chain Rule.
  
  
  • $\dfrac{d}{d𝒙}\big[\sqrt{\,𝒙^{3}+\csc(𝒙^{3})\,}\big]=\dfrac{d}{d𝒙}\big[(𝒙^{3}+\csc(𝒙^{3}))^{1/2}\big]$
  
  
  • $\dfrac{1}{2}(𝒙^{3}+\csc(𝒙^{3}))^{-1/2}\cdot\dfrac{d}{dx}\big[𝒙^{3}+\csc(𝒙^{3})\big]$
  
  
  • $\dfrac{1}{2}(𝒙^{3}+\csc(𝒙^{3}))^{-1/2}\cdot\big[3𝒙^{2}+\dfrac{d}{dx}\big(\csc(𝒙^{3})\big)\big]$
  
  
  • $\dfrac{1}{2}(𝒙^{3}+\csc(𝒙^{3}))^{-1/2}\cdot\big[3𝒙^{2}+(-\csc(𝒙^{3})\cot(𝒙^{3}))\cdot\dfrac{d}{dx}[𝒙^{3}]\big]$
  
  
  • $\dfrac{1}{2}(𝒙^{3}+\csc(𝒙^{3}))^{-1/2}\cdot\big[3𝒙^{2}-3𝒙^{2}\csc(𝒙^{3})\cot(𝒙^{3})\big]$
  
  
  • $\dfrac{\,3𝒙^{2}-3𝒙^{2}\csc(𝒙^{3})\cot(𝒙^{3})\,}{\,2\sqrt{\,𝒙^{3}+\csc(𝒙^{3})\,}}$
  
  
• 1:19:15. 🧩 Example –: $\dfrac{d}{d𝒙}\big[(3+𝒙^{2}\cdot\cot(𝒙^{2}))^{-3}\big]$ – [📷image]
  • Example illustrating the convenience of simplifying expressions before applying the Quotient Rule.
  • In this case, $\dfrac{d}{dx}\Big[\dfrac{1}{\big(3+𝒙^{2}\cdot\cot(𝒙^{2})\big)^{3}}\Big]$, it is recommended to move the denominator to the numerator with a negative exponent to apply the Generalized Power Rule.
  • Avoid using the Quotient Rule, as it can complicate the differentiation process.
  
  
  • $\dfrac{d}{dx}\big[3+𝒙^{2}\cot(𝒙^{2})\big]^{-3}=-3\big[3+𝒙^{2}\cot(𝒙^{2})\big]^{-4}\cdot\dfrac{d}{dx}\big[3+𝒙^{2}\cot(𝒙^{2})\big]$
  
  
  • $-3\big[3+𝒙^{2}\cot(𝒙^{2})\big]^{-4}\cdot\big[\dfrac{d}{dx}(𝒙^{2})\cdot\cot(𝒙^{2})+𝒙^{2}\cdot\dfrac{d}{dx}(\cot(𝒙^{2}))\big]$
  
  
  • $-3\big[3+𝒙^{2}\cot(𝒙^{2})\big]^{-4}\cdot\big[2𝒙\cdot\cot(𝒙^{2})+𝒙^{2}\cdot(-\csc^{2}(𝒙^{2})\cdot\dfrac{d}{dx}[𝒙^{2}])\big]$
  
  
  • $=-3\big[3+𝒙^{2}\cot(𝒙^{2})\big]^{-4}\cdot\big[2𝒙\cdot\cot(𝒙^{2})+𝒙^{2}\cdot(-\csc^{2}(𝒙^{2})\cdot 2𝒙)\big]$
  
  
  • $-3\big[3+𝒙^{2}\cot(𝒙^{2})\big]^{-4}\cdot\big[2𝒙\cot(𝒙^{2})-2𝒙^{3}\csc^{2}(𝒙^{2})\big]$
  
  
  • $-\dfrac{3\big[2𝒙\cot(𝒙^{2})-2𝒙^{3}\csc^{2}(𝒙^{2})\big]}{\big[3+𝒙^{2}\cot(𝒙^{2})\big]^{4}}$
  
  
• 1:31:20. 🧩 Example –: $\dfrac{d}{d𝒙}\Big[\dfrac{1+\cos(𝒙^{2})}{1+\sin(𝒙^{2})}\Big]$ – [📷image]
  • If $\dfrac{d}{d𝒙}\Big[\dfrac{1+\cos(𝒙^{2})}{1+\sin(𝒙^{2})}\Big]^{5}$ then apply general power rule, quotient rule and chain rule
  • Apply the **quotient rule**:
    
    
    • $\dfrac{d}{dx}\Big[\dfrac{𝒇(𝒙)}{𝗀(𝒙)}\Big]=\dfrac{𝒇'(𝒙)\cdot 𝗀(𝒙)-𝒇(𝒙)\cdot 𝗀'(𝒙)}{\big(𝗀(𝒙)\big)^{2}}$
      
      
      • Here: $𝒇(𝒙)=1+\cos(𝒙^{2})$ aand $𝗀(𝒙)=1+\sin(𝒙^{2})$
      
      
    • Compute the derivatives using the **chain rule**:
      • $𝒇'(𝒙)=\dfrac{d}{dx}\big[1+\cos(𝒙^{2})\big]=-\sin(𝒙^{2})\cdot 2𝒙$
      
      
      • $𝗀'(𝒙)=\dfrac{d}{dx}\big[1+\sin(𝒙^{2})\big]=\cos(𝒙^{2})\cdot 2𝒙$
      
      
    • Plug into the quotient rule formula:
      • Numerator:
        • $𝒇'(𝒙)\cdot 𝗀(𝒙)-𝒇(𝒙)\cdot 𝗀'(𝒙)$
        • $\big[-\sin(𝒙^{2})\cdot 2𝒙\big]\cdot\big[1+\sin(𝒙^{2})\big]-\big[1+\cos(𝒙^{2})\big]\cdot\big[\cos(𝒙^{2})\cdot 2𝒙\big]$
        • $2𝒙\big[-\sin(𝒙^{2})-\sin^{2}(𝒙^{2})-\cos(𝒙^{2})-\cos^{2}(𝒙^{2})\big]$
      • Denominator:
        • $\big[1+\sin(𝒙^{2})\big]^{2}$
    • Final answer:
      • $\dfrac{d}{dx}\Big[\dfrac{1+\cos(𝒙^{2})}{1+\sin(𝒙^{2})}\Big]$ $=\dfrac{2𝒙\big[-\sin(𝒙^{2})-\sin^{2}(𝒙^{2})-\cos(𝒙^{2})-\cos^{2}(𝒙^{2})\big]}{\big[1+\sin(𝒙^{2})\big]^{2}}$
• 1:32:10. 🧩 Example –: $\dfrac{d}{d𝒙}\big[𝒙^{2}\cdot\sin(3𝒙)\big]$ – [📷image]
  • Product rule and then chain rule
  • Apply the product rule:
    • $\dfrac{d}{dx}\big[𝒇(𝒙)\cdot 𝗀(𝒙)\big]=𝒇'(𝒙)\cdot 𝗀(𝒙)+𝒇(𝒙)\cdot 𝗀'(𝒙)$
      • Here $𝒇(𝒙)=𝒙^{2}$ and $𝗀(𝒙)=\sin(3𝒙)$
    • Compute the derivatives:
      • $𝒇'(𝒙)=2𝒙$
      • $𝗀'(𝒙)=\dfrac{d}{dx}\big[\sin(3𝒙)\big]=\cos(3𝒙)\cdot 3$ (by the **chain rule**)  
    • Now substitute into the product rule formula:
      • $\dfrac{d}{dx}\big[𝒙^{2}\cdot\sin(3𝒙)\big]=2𝒙\cdot\sin(3𝒙)+𝒙^{2}\cdot\big[\cos(3𝒙)\cdot 3\big]$
      • $2𝒙\cdot\sin(3𝒙)+3𝒙^{2}\cdot\cos(3𝒙)$